Question

for each balanced reaction,

calculate the mass of each product formed if 14.5 g of the first reactant is combined with 9.40 g of the second reactant.

1) 2Al(s) + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2(g)

2) 2AgNO₃(aq) + CaSO₄(aq) → Ag₂SO₄(s) + Ca(NO₃)₂(aq)

3) CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l)

Answer 1

1) 14.5 g of Al = 14.5 / 27 = 0.537 moles

9.40 g of HNO3 = 9.40 / 63 = 0.149 moles

HNO3 is the limiting reagent

So the products produced will be :

0.049 moles of Al(NO3)3 = 0.049 x 212.99 = 10.59 g

0.074 moles of H2 = 0.074 x 2.016 = 0.15 g

2) 14.5 g of AgNO3 = 14.5 / 169.8 = 0.085 moles

9.40 g of CaSO4 = 9.40 / 136.1 = 0.069 moles

AgNO3 is the limiting reagent

So the products :

0.043 moles of Ag2SO4 = 0.043 x 311.799 = 13.3 g

0.043 moles of Ca(NO3)2 = 0.043 x 164.09 = 7.003 g

3) 14.5 g of CaO = 14.5 / 56 = 0.26 moles

9.40 g of HCl = 9.40 / 36.46 = 0.25 moles

HCl is the limiting reagent

products formed will be :

0.129 moles of CaCl2 = 0.129 x 110.98 = 14.31 g

0.129 moles of H2O = 0.129 x 18.01528 = 2.32 g