5g of calcium carbonate was strongly heated to a constant mass calculate the mas of the solid residue formed Ca=40.0,O=16.0,C=12
Solution:
The molar mass of calcium carbonate (CaCO3) is (40 + 12 + 3×16) = 100 g/mol
Therefore,
Moles of CaCO3 = (5 g CaCO3) × (1 mol CaCO3 / 100 g CaCO3) = 0.05 mol CaCO3
Balanced chemical equation:
CaCO3(s) → CaO(s) + CO2(g)
Thus, the solid residue is calcium oxide (CaO)
According to stoichiometry:
1 mol of CaCO3 produces 1 mol of CaO
Thus, 0.05 mol of CaCO3 produce:
(0.05 mol CaCO3) × (1 mol CaO / 1 mol CaCO3) = 0.05 mol CaO
The molar mass of calcium oxide (CaO) is (40 + 16) = 56 g/mol
Therefore,
Mass of CaO = (0.05 mol CaO) × (56 g CaO / 1 mol CaO) = 2.8 g CaO
Mass of CaO = 2.8 g
Answer: The mas of the solid residue (CaO) is 2.8 grams
Comments
Leave a comment