Question

if 0.615 g of magnesium hydroxide reacts with 0.810 g sulfuric acid, what is the mass of the magnesium sulfate produced ?

Answer 1

Balanced equation:

Mg(OH)2(s) + H2SO4(l) ====> MgSO4(s) + 2 H2O(l)

0.619 g of magnesium hydroxide = 0.619/58.31 = 0.01061 Moles

.0.940 g of sulfuric acid = 0.94 / 98.07 = 0.009584 Moles

moles of magnesium sulfate produced = 0.009584 Moles

mass of magnesium sulfate produced = 0.009584 x 120.36 = 1.153 gm