Answer
$O_{2}$ is the limiting reactant.
Work Step by Step
1. Find the nº of moles in each reagent.
nº of moles = mass (g) / molar mass
$C_{2}H_{2}$: 12*2 + 1*2 = 26 g/mol
$O_{2}$: 16*2 = 32 g/mol
$C_{2}H_{2}$:nº of moles = 10 g $\div$26 g/mol = 0.384 mol
$O_{2}$: nº of moles = 10 g $\div$ 32 g/mol = 0.3122.5 mol
2. Divide the nº of moles by the coefficient in the equation.
$C_{2}H_{2}$: 0.384$\div$2 = 0.192
$O_{2}$: 0.3125$\div$5 = 0.0625
3. Because the $O_{2}$ value is the lowest, it is the limiting reagent.
