Answer to Question #112143 in General Chemistry for Taylor

Question

2HCl + Mg → MgCl2 + H2

If we use 12grams of HCl and 8grams of Mg, which is the limiting reactant, and which reactant is excess?

Answer 1

n(HCl) = m/M = 12g/36.5g/mol = 0.329 mol

n(Mg) = 8g/24g/mol = 0.333 mol

According to the equation 1 mol of Mg reacts with 2 mol of HCl.

HCl is the limiting reactant, Mg is in the excess

n(Mg) = 0.333/2 = 0.1665 mol reacts

n(Mg) = 0. 329 - 0.1665 = 0.1625 mol in the excess