Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Given 10 g of Zn and 10 g HCl; what volume of Hy is produced?

3.02 L of will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

According to stoichiometry :

2 moles of require  =  1 mole of

Thus 0.27 moles of will require= of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 2 moles of give = 1 moles of

Thus 0.27 moles of give =  of

Volume of

Thus 3.02 L of will be produced from the given masses of both reactants.