Question

Magnesium carbonate (MgCO3) decomposes as shown below.
MgCO3(s)  MgO(s) + CO2(g)
How many grams of magnesium oxide will be formed when 84 grams of magnesium carbonate decomposes completely?
Given the atomic masses:
Mg =24, C = 12, and O = 16

Answer 1

m(MgCO3) = 84 g
M(MgCO3) = 24 + 12 + 3*16 = 84 g/mol
n(MgCO3) = m/M = 84g/84g/mol = 1 mol
n(MgCO3) = n(MgO) = 1 mol
M(MgO) = 24 + 16 = 40 g/mol
m(MgO) = n*M = 1mol*40 g/mol = 40 g