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Question
question 25 of 37
balance the following chemical equation (if necessary):
c₆h₆(l) + o₂(g) → h₂o(g) + co₂(g)
Step1: Balance carbon atoms
There are 6 carbon atoms in \(C_6H_6\). So, we put a coefficient of 6 in front of \(CO_2\).
\(C_6H_6(l)+O_2(g)\to H_2O(g)+6CO_2(g)\)
Step2: Balance hydrogen atoms
There are 6 hydrogen atoms in \(C_6H_6\). So, we put a coefficient of 3 in front of \(H_2O\).
\(C_6H_6(l)+O_2(g)\to 3H_2O(g)+6CO_2(g)\)
Step3: Balance oxygen atoms
On the right - hand side, the number of oxygen atoms in \(3H_2O\) is \(3\times1 = 3\) and in \(6CO_2\) is \(6\times2=12\). So, the total number of oxygen atoms on the right - hand side is \(3 + 12=15\). On the left - hand side, we have \(O_2\). Let the coefficient of \(O_2\) be \(x\). Then \(2x = 15\), \(x=\frac{15}{2}\). But we usually prefer whole - number coefficients. So, we multiply the entire equation by 2.
Multiplying the equation \(C_6H_6(l)+O_2(g)\to 3H_2O(g)+6CO_2(g)\) by 2 gives:
\(2C_6H_6(l)+15O_2(g)\to 6H_2O(g)+12CO_2(g)\)
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\(2C_6H_6(l)+15O_2(g)\to 6H_2O(g)+12CO_2(g)\)