Question

For each of the reactions, calculate the mass (in grams) of the product formed when 3.75 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

Ba(s)−−−−−+Cl2(g)→BaCl2(s)

CaO(s)−−−−−−+CO2(g)→CaCO3(s)

2Mg(s)−−−−−+O2(g)→2MgO(s)

Answer 1

1)

5.687g

Explanation

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

Stoichiometrically, 1mole of Ba gives 1mole of BaCl2

No of moles = mass/molar mass

given moles of Ba = 3.75g/137.327g/mol = 0.02731mol

0.02731moles of Ba should give 0.02731 moles of BaCl2

Molar mass of BaCl2 = 208.23g/mol

mass of BaCl2 formed = 0.02731mol × 208.23g/mol = 5.687g

2)

6.6923g

Explanation

CaO(s) + CO2(g) -----> CaCO3(s)

stoichiometrically, 1mole of CaO gives 1mole of CaCO3

given moles of CaO = 3.75g /56.078g/mol = 0.06687mol

0.06687moles of CaO should give 0.06687moles of CaCO3

Molar mass of CaCO3 = 100.08g/mol

mass of CaCO3 formed = 0.06687mol × 100.08g/mol = 6.6923g

3)

6.2187g

Explanation

2Mg(s) + O2(g) ------> 2MgO(s)

Stoichiometrically, 2moles of Mg gives 2moles of MgO

given moles of Mg =3.75g/ 24.305g/mol = 0.15429mol

0.15429 moles of Mg should give 0.15429moles of MgO

molar mass of MgO = 40.305g/mol

mass of MgO formed = 0.15429mol × 40.305g/mol = 6.2187g