Zn + 2 HCl → ZnCl2 + H2 If 381g of HCl are reacted with 307g of Zn, how many grams of H2 will be produced?

10.1g of H₂ are produced

Explanation:

To solve this question we need, first, to convert the mass of each reactant to moles and, using the chemical reaction, find limiting reactant. With limiting reactant we can find the moles of H2 and its mass:

Moles Zn -Molar mass: 65.38g/mol-:

307g * (1mol / 65.38g) = 4.696 moles

Moles HCl -Molar mass: 36.46g/mol-:

381g HCl * (1mol / 36.46g) = 10.45 moles

For a complete reaction of 10.45 moles of HCl are required:

10.45 moles HCl * (1mol Zn / 2mol HCl) = 5.22 moles Zn

As there are 4.696 moles of Zn, Zn is the limiting reactant

The moles of H₂ produced = Moles of Zn added = 4.696 moles. The mass is-Molar mass H₂ = 2.16g/mol-:

4.696 moles * (2.16g / mol) =