Question

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
How many moles of H2O are produced when 64.0 g C2H2 burn in oxygen?

M.M

H2O = 18.02 g/mol

C2H2 = 26.04 g/mol


2.46 moles H2O
5.13 moles H2O
4.92 moles H2O
2.00 moles H2O

Answer 1

Answer:

Approximately $2.46\; \rm mol$.

Explanation:

Make use of the molar mass data ($M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$:

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$.

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$.

Coefficient of $\rm H_2O$ in this equation: $2$.

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$.

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$.