Answer:
16.02 g
Explanation:
the balanced equation for the decomposition of CuCO₃ is as follows
CuCO₃ --> CuO + CO₂
molar ratio of CuCO₃ to CO₂ is 1:1
number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol
according to the molar ratio
1 mol of CuCO₃ decomposes to form 1 mol of CO₂
therefore 0.364 mol of CuCO₃ decomposes to form 0.364 mol of CO₂
number of CO₂ moles produced - 0.364 mol
therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g
16.02 g of CO₂ produced
Answer:
Percent composition tells you which types of atoms (elements) are present in a molecule and their levels. Percent composition can also tell you about the different elements present in an ionic compound as well.
