Question

For each of the reactions, calculate the mass (in grams) of the product formed when 3.33 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

1. Ba(s)+Cl2(g)→BaCl2(s)

2. CaO(s)+CO2(g)→CaCO3(s)

3. 2Mg(s)+O2(g)→2MgO(s)

4. 4Al(s)+3O2(g)→2Al2O3(s)

Answer 1

1) Ba(s) + Cl2(g) -----> BaCl2(s)

Cl2 is in excess, then moles of Ba = 3.33g / 137.32g/mol = 0.02425 moles

1 mole of Ba gives 1 mole of BaCl2

then mass of 0.02425 moles of BaCl2 = 0.02425 mol * 208.23 g/mol = 5.05 g of BaCl2

2) CaOs) + CO2(g) -----> CaCO3(s)

CO2 is in excess, then moles of CaO = 3.33g / 56.08g/mol = 0.05938 moles

1 mole of CaO gives 1 mole of CaCO3

then mass of 0.05938 moles of CaCO3 = 0.05938 mol * 100.09g/mol = 5.94 g of CaCO3

3)  2Mg(s)+O2(g)→2MgO(s)

O2 is in excess, then moles of Mg = 3.33g / 24.31g/mol = 0.1370moles

1 mole of Mg gives 1 mole of MgO

then mass of 0.1370 moles of MgO = 0.1370 mol * 40.30 g/mol = 5.52 g of MgO

4) 4Al(s)+3O2(g)→2Al2O3(s)

O2 is in excess, then moles of Al = 3.33g / 26.98 g/mol = 0.1234 moles

1 mole of Al gives 2 mole of Al2O3

then mass of 0.06171 moles of Al2O3 = 0.06171 mol * 101.96 g/mol = 6.29 g of Al2O3