Answer to Question #102042 in General Chemistry for Shikha Patel

Question #102042

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

Expert's answer

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

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