## Chemistry: The Central Science (13th Edition)

$O_{2}$ is the limiting reactant.
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.