Question

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?

Answer 1

`0.8703"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"g"_"Na" * (1mol_"Na")/(22.99"g"_"Na") * (1"rxn")/(2"mol"_"Na") = 0.6901"rxn"`

`31.73"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"g"_"HCl" * (1mol_"HCl")/(36.46g_"HCl") * (2"mol"H2)/(2"mol"_"HCl") = 0.8703"mol"H_2`