# How many mols of H2 can be produced from the equation ( 2Na+2HCl yield 2NaCl+H2) if you begin with 31.73grams of each reactant?

Apr 26, 2018

$0.8703 \text{mol} {H}_{2}$

#### Explanation:

Before we can find the mols of ${H}_{2}$, we need to find the limiting reactant. We can find this by multiplying the mols of our reactants by the mols needed for a reaction.

Note: rxn = Reaction

$31.73 \text{g"_"Na" * (1mol_"Na")/(22.99"g"_"Na") * (1"rxn")/(2"mol"_"Na") = 0.6901"rxn}$

$31.73 \text{g"_"HCl" * (1mol_"HCl")/(36.46"g"_"HCl") * (1"rxn")/(2mol_"HCl") = 0.4351"rxn}$

Since the Hydrogen Chloride has less reactions, it is the limiting reactant, so we should use this in our calculations.

$31.73 \text{g"_"HCl" * (1mol_"HCl")/(36.46g_"HCl") * (2"mol"H2)/(2"mol"_"HCl") = 0.8703"mol} {H}_{2}$