Question

For each of the reactions, calculate the mass (in grams) of the product formed when 3.26g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

Part A Ba(s)?????+Cl2(g)?BaCl2(s) m=_________________g

Part B CaO(s)??????+CO2(g)?CaCO3(s) m=_________________g

Part C 2Mg(s)?????+O2(g)?2MgO(s) m=_________________g

Part D 4Al(s)????+3O2(g)?2Al2O3(s) m=_________________g

Answer 1

The reactant in lesser amount is the limiting reagent . from the No of moles of limiting reagent we can calculate the No of moles of product. and then its mass

Part A

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

mass of Ba(s)=3.26g

No of moles of Ba(s)=mass/molar mass=3.26g/137.33g/mol=0.02374mol

from the above equation if 1 mole of Ba(s) reacts, one mole of BaCl will form

so 0.02374mol of Ba(s) reacts to give 0.02374mol of BaCl

mass of BaCl=No of molesxmolar mass=0.02374molx208.23 g/mol=4.9430g

Part B

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

mass of CaO(s)=3.26g

No of moles of CaO(s)=mass/molar mass=3.26g/56.0774 g/mol=0.05813mol

from the above equation if 1 mole of CaO((s) reacts, one mole of CaCO3 will form

so 0.05813mol of CaO((s) reacts to give 0.05813mol of CaCO3

mass of CaCO3 =No of molesxmolar mass=0.05813molx100.0869 g/mol=5.8184g

Part C

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

mass of Mg(s)=3.26g

No of moles of Mg(s)=mass/molar mass=3.26g/24.305g/mol=0.1341mol

from the above equation if 2 moles of Mg(s) reacts, 2 moles of MgO(s) will form

so 0.1341mol of Mg(s) reacts to give 0.1341mol of MgO(s)

mass of MgO(s)=No of molesxmolar mass=0.1341molx40.3044 g/mol=5.4060g

Part D

4Al(s)+3O2(g)------>2Al2O3(s)

mass of Al(s)=3.26g

No of moles of Al(s)=mass/molar mass=3.26g/26.98g/mol=0.1208mol

from the above equation if 4 moles of Al(s) reacts, 2 moles of Al2O3(s) will form

so 0.1208mol of Al(s) reacts to give moles of Al2O3(s)=2molx0.1208mol/4mol=0.0604mol

mass of Al2O3(s)=No of molesxmolar mass=0.0604molx101.96 g/mol=6.1599g