How many moles of oxygen gas are required to combust a 4*mol quantity of acetylene completely?

Mar 11, 2016

$H - C \equiv C - H \left(g\right) + \frac{5}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + {H}_{2} O \left(g\right)$

Explanation:

The balanced chemical equation is given. Each mole of acetylene requires two and a half moles of oxygen gas. So if 4 mol acetylene are combusted, some 10 moles of dioxygen are necessary.

Mar 11, 2016

$10$ $m o l$ ${O}_{2}$

Explanation:

1. First thing to do is to balance the chemical equation; The balanced equation is:
$2 {C}_{2} {H}_{2} + 5 {O}_{2} \rightarrow 4 C {O}_{2} + 2 {H}_{2} O$

2. Then use the coefficients, which represents the number of mol of its compound/molecules in said reaction, to get the number of mol of ${O}_{2}$ molecules;

3. Multiply the mol of acetylene as provided in the problem ($4 m o l {C}_{2} {H}_{2}$) with the convertion ratio of the moles of acetylene and oxygen molecules ((5 mol O_2)/(2mol C_2H_2)) based on the balanced equation;
4. Make sure to cancel the mole of ${C}_{2} {H}_{2}$ and get your desired mole of ${O}_{2}$ molecules.
5. The answer per computation is 10 mol of ${O}_{2}$moleules