Question

For each of the reactions, calculate the mass (in grams) of the product formed when 3.30 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

(underlined) Ba(s)?????+Cl2(g)?BaCl2(s)

(underlined) CaO(s)??????+CO2(g)?CaCO3(s)

(underlined) 2Mg(s)?????+O2(g)?2MgO(s)  

(underlined) 4Al(s)?????+3O2(g)?2Al2O3(s)

Answer 1

Ba(s)+Cl2(g) →BaCl2(s)

No of mol of Ba   = 3.3/137.327 = 0.024 mol

mass of BaCl2 = 0.024*208.2330 = 5 grams

CaO(s)+CO2(g) → CaCO3(s)

No of mol of ca   = 3.3/40 = 0.0825 mol

mass of caco3 = 0.0825*100.0869   = 8.25 grams

2Mg(s)+O2(g) → 2MgO(s)

No of mol of Mg   = 3.3/24 = 0.1375 mol

mass of MgO = 0.1375*40.3044   = 5.54 grams

4Al(s) +3O2(g)→2Al2O3(s)

No of mol of aL   = 3.3/27 = 0.12 mol

mass of MgO = 0.12*(2/4)*101.96   = 6.12 grams