Question

2Mg(s)+ O₂(g) ----> 2MgO(s)

a) What is the theoretical yield of MgO(s), in moles, when 6.50g Mg(s) reacts with 3.45g of O₂(g)

b) What was the limiting reactant(LR) and the mass (in g) of MgO formed?

Answer 1

molar mass of Mg = 24 g/mol
molar mass of O2 = 32 g/mol
number of mole = (given mass)/(molar mass)
number of mole of Mg = 6.50/24
= 0.27 mole
number of mole of O2 = 3.45/32
= 0.11 mole

reaction taking place is
2Mg + O2 --> MgO
according to reaction
2 mol of Mg required 1 mol of O2
0.27 mol of Mg required 0.135 mol of O2
but we have only 0.11 mole of O2
so, O2 is limiting reagent

a)
again
1 mol of O2 give 2 mol of MgO
0.11 mol of O2 give 0.22 mol of MgO
so,
number of mole of MgO formed = 0.22 mole

b)
here, limiting reagent is O2
molar mass of MgO = 40 g/mol
mass of MgO formed = (number of mole of MgO formed)*(molar mass of MgO)
= 0.22*40
= 8.8 g