Question

If 32.00 g of MgCO3 decomposes in this way, the mass of solid product expected is _______.

Answer 1

MgCO3 = MgO +CO2

CO2 - gas

MgO - solid

m_MgO = n_MgO * M_MgO

M_MgO= 40 g/mol

n_MgO = n_MgCO3 = 32/ 84 = 0.38 mol

m_MgO = 0.38* 40 = 15.24 g