Question

question 24 of 37
balance the following chemical equation (if necessary):
si₂h₃(s) + o₂(g) → sio₂(g) + h₂o(g)

Explanation:

Step1: Balance the Si atoms

There are 2 Si atoms on the left - hand side (in \(Si_2H_3\)). So, we put a coefficient of 2 in front of \(SiO_2\) on the right - hand side.
\(Si_2H_3(s)+O_2(g)\to2SiO_2(g)+H_2O(g)\)

Step2: Balance the H atoms

There are 3 H atoms on the left - hand side (in \(Si_2H_3\)). To balance H, we put a coefficient of \(\frac{3}{2}\) in front of \(H_2O\) on the right - hand side.
\(Si_2H_3(s)+O_2(g)\to2SiO_2(g)+\frac{3}{2}H_2O(g)\)

Step3: Balance the O atoms

On the right - hand side, the number of O atoms in \(2SiO_2\) is \(2\times2 = 4\) and in \(\frac{3}{2}H_2O\) is \(\frac{3}{2}\). So the total number of O atoms on the right - hand side is \(4+\frac{3}{2}=\frac{8 + 3}{2}=\frac{11}{2}\). To balance O, we put a coefficient of \(\frac{11}{4}\) in front of \(O_2\) on the left - hand side.
\(Si_2H_3(s)+\frac{11}{4}O_2(g)\to2SiO_2(g)+\frac{3}{2}H_2O(g)\)

Step4: Eliminate fractions

Multiply each coefficient by 4 to get rid of the fractions.
\(4Si_2H_3(s)+11O_2(g)\to8SiO_2(g)+6H_2O(g)\)

Answer:

\(4Si_2H_3(s)+11O_2(g)\to8SiO_2(g)+6H_2O(g)\)