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

Approximately
`2.46\; \rm mol`.

Step-by-step 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} \\ &= (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})} = (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 * 1 = 2.46\; \rm mol\end{aligned}`.