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 Nov 19 at 7:45
Arthur
109k7103186
asked Nov 19 at 7:42
Mit34
1
add a comment |
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 Nov 19 at 7:45
Arthur
109k7103186
asked Nov 19 at 7:42
Mit34
1
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
Nov 19 at 7:47
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
Nov 19 at 7:48
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 Nov 19 at 7:45
Arthur
109k7103186
asked Nov 19 at 7:42
Mit34
1
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 Nov 19 at 7:45
Arthur
109k7103186
asked Nov 19 at 7:42
Mit34
1
edited Nov 19 at 7:45
Arthur
109k7103186
asked Nov 19 at 7:42
Mit34
1
edited Nov 19 at 7:45
Arthur
109k7103186
edited Nov 19 at 7:45
Arthur
109k7103186
edited Nov 19 at 7:45
Arthur
109k7103186
109k7103186
asked Nov 19 at 7:42
Mit34
1
asked Nov 19 at 7:42
Mit34
1
asked Nov 19 at 7:42
Mit34
1
1
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
Nov 19 at 7:47
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
Nov 19 at 7:48
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
Nov 19 at 7:47
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
Nov 19 at 7:48
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
Nov 19 at 7:47
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
Nov 19 at 7:47
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
Nov 19 at 7:48
@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
Nov 19 at 7:48
add a comment |
1 Answer
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1
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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 Nov 19 at 7:50
Arthur
109k7103186
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 Nov 19 at 7:50
Arthur
109k7103186
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 Nov 19 at 7:50
Arthur
109k7103186
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 Nov 19 at 7:50
Arthur
109k7103186
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 Nov 19 at 7:50
Arthur
109k7103186
edited Nov 19 at 7:59
answered Nov 19 at 7:50
Arthur
109k7103186
answered Nov 19 at 7:50
Arthur
109k7103186
answered Nov 19 at 7:50
Arthur
109k7103186
109k7103186
add a comment |
add a comment |
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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
Nov 19 at 7:47
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
Nov 19 at 7:48