Mixing
Writing a balanced chemical equation with linear systems
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Could someone please help explain how they got to this next step (writing in tabular form) in solving this chemical equation? (I have the worked answer, but I don't understand the first step they did).
linear-algebra
edited 5 hours ago
Arthur
107k7103186
asked 5 hours ago
Mit34
1
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Could someone please help explain how they got to this next step (writing in tabular form) in solving this chemical equation? (I have the worked answer, but I don't understand the first step they did).
linear-algebra
edited 5 hours ago
Arthur
107k7103186
asked 5 hours ago
Mit34
1
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago
1
@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though
– postmortes
5 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Could someone please help explain how they got to this next step (writing in tabular form) in solving this chemical equation? (I have the worked answer, but I don't understand the first step they did).
linear-algebra
edited 5 hours ago
Arthur
107k7103186
asked 5 hours ago
Mit34
1
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could someone please help explain how they got to this next step (writing in tabular form) in solving this chemical equation? (I have the worked answer, but I don't understand the first step they did).
linear-algebra
linear-algebra
edited 5 hours ago
Arthur
107k7103186
asked 5 hours ago
Mit34
1
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Arthur
107k7103186
asked 5 hours ago
Mit34
1
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Arthur
107k7103186
edited 5 hours ago
Arthur
107k7103186
edited 5 hours ago
Arthur
107k7103186
107k7103186
asked 5 hours ago
Mit34
1
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Mit34
1
asked 5 hours ago
Mit34
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1
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago
1
@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though
– postmortes
5 hours ago
add a comment |
Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago
1
@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though
– postmortes
5 hours ago
Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago
Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago
1
1
@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though
– postmortes
5 hours ago
@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though
– postmortes
5 hours ago
add a comment |
1 Answer
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up vote
1
down vote
If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
$$
x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
$$
has as many of each atom on the left side as it does on the right side.
Looking at carbon first, the number of carbon atoms on the left side is $6x_1$, while on the right side it's $x_2+2x_3$. These two numbers are supposed to be equal, so we put $=$ between them, and that's the first equation.
The two other equations are done similarly for hydrogen and oxygen. However, there is a typo in your picture: it's supposed to be $x_1$ on the left side of all three equations.
That is, after all, how many sugar molecules there are on the left-hand side. Thus we get
$$
begin{array}{lccc}
text{Element} &text{Left side} &&text{Right side}\
text{Carbon}&6x_1&=&x_2+2x_3\
text{Hydrogen} &12x_1&=&6x_3\
text{Oxygen} &6x_1&=&2x_2+x_3
end{array}
$$
answered 5 hours ago
Arthur
107k7103186
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
$$
x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
$$
has as many of each atom on the left side as it does on the right side.
Looking at carbon first, the number of carbon atoms on the left side is $6x_1$, while on the right side it's $x_2+2x_3$. These two numbers are supposed to be equal, so we put $=$ between them, and that's the first equation.
The two other equations are done similarly for hydrogen and oxygen. However, there is a typo in your picture: it's supposed to be $x_1$ on the left side of all three equations.
That is, after all, how many sugar molecules there are on the left-hand side. Thus we get
$$
begin{array}{lccc}
text{Element} &text{Left side} &&text{Right side}\
text{Carbon}&6x_1&=&x_2+2x_3\
text{Hydrogen} &12x_1&=&6x_3\
text{Oxygen} &6x_1&=&2x_2+x_3
end{array}
$$
answered 5 hours ago
Arthur
107k7103186
add a comment |
up vote
1
down vote
If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
$$
x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
$$
has as many of each atom on the left side as it does on the right side.
Looking at carbon first, the number of carbon atoms on the left side is $6x_1$, while on the right side it's $x_2+2x_3$. These two numbers are supposed to be equal, so we put $=$ between them, and that's the first equation.
The two other equations are done similarly for hydrogen and oxygen. However, there is a typo in your picture: it's supposed to be $x_1$ on the left side of all three equations.
That is, after all, how many sugar molecules there are on the left-hand side. Thus we get
$$
begin{array}{lccc}
text{Element} &text{Left side} &&text{Right side}\
text{Carbon}&6x_1&=&x_2+2x_3\
text{Hydrogen} &12x_1&=&6x_3\
text{Oxygen} &6x_1&=&2x_2+x_3
end{array}
$$
answered 5 hours ago
Arthur
107k7103186
add a comment |
up vote
1
down vote
up vote
1
down vote
If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
$$
x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
$$
has as many of each atom on the left side as it does on the right side.
Looking at carbon first, the number of carbon atoms on the left side is $6x_1$, while on the right side it's $x_2+2x_3$. These two numbers are supposed to be equal, so we put $=$ between them, and that's the first equation.
The two other equations are done similarly for hydrogen and oxygen. However, there is a typo in your picture: it's supposed to be $x_1$ on the left side of all three equations.
That is, after all, how many sugar molecules there are on the left-hand side. Thus we get
$$
begin{array}{lccc}
text{Element} &text{Left side} &&text{Right side}\
text{Carbon}&6x_1&=&x_2+2x_3\
text{Hydrogen} &12x_1&=&6x_3\
text{Oxygen} &6x_1&=&2x_2+x_3
end{array}
$$
answered 5 hours ago
Arthur
107k7103186
If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
$$
x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
$$
has as many of each atom on the left side as it does on the right side.
Looking at carbon first, the number of carbon atoms on the left side is $6x_1$, while on the right side it's $x_2+2x_3$. These two numbers are supposed to be equal, so we put $=$ between them, and that's the first equation.
The two other equations are done similarly for hydrogen and oxygen. However, there is a typo in your picture: it's supposed to be $x_1$ on the left side of all three equations.
That is, after all, how many sugar molecules there are on the left-hand side. Thus we get
$$
begin{array}{lccc}
text{Element} &text{Left side} &&text{Right side}\
text{Carbon}&6x_1&=&x_2+2x_3\
text{Hydrogen} &12x_1&=&6x_3\
text{Oxygen} &6x_1&=&2x_2+x_3
end{array}
$$
answered 5 hours ago
Arthur
107k7103186
edited 5 hours ago
answered 5 hours ago
Arthur
107k7103186
answered 5 hours ago
Arthur
107k7103186
answered 5 hours ago
Arthur
107k7103186
107k7103186
add a comment |
add a comment |
Mit34 is a new contributor. Be nice, and check out our Code of Conduct.
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Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago
1
@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though
– postmortes
5 hours ago