calculate mass of phosphoric acid obtained from 28.4 g of phosphorus pentoxide. solve by POAC
Answers
Answer:
1. STOICHIOMETRY INVOLVING ONLY PURE SUBSTANCES
For all chemical reactions, the balanced chemical equation gives the mole ratios of reactants and
products. If we are dealing with pure chemicals, the molar mass allows us to convert the mass of a
reactant or product into moles. Consider the reaction shown below.
Ca3N2(s) + 6H2O(l) H 2NH3(g) + 3Ca(OH)2(s)
mole ratio 1 : 6 : 2 : 3
molar mass 148.3 18.0 17.0 74.1
(g/mole)
For this reaction, 1 mole of Ca3N2 will react with 6 moles of H2O to produce 2 moles of NH3 and 3
moles of Ca(OH)2. Therefore, for this reaction 1 mole Ca3N2 ≡ 6 moles H2O. Similar equivalences will
apply to other pairs of reactants and/or products. For example, 6 moles H2O ≡ 2 moles NH3. As discussed
before, each equivalence will give two conversion factors.
(a) Calculate the number of moles of H2O that will react with 2.5 moles of Ca3N2.
Solution: moles H2O = = 15 moles H2O
6 moles H2O
1 mole Ca3N2
2.5 moles Ca3N2 x
given mole ratio
(b) How many moles of NH3 can be made by the reaction of 0.75 mole of Ca3N2 with an excess of
H2O? (Ans. 1.5 mole)
(c) How many moles of H2O are needed to make 12 moles of NH3? (Ans. 36 moles)
(d) Calculate the mass of Ca(OH)2 that can be formed from the reaction of 10.0 g of Ca3N2 with an
excess of H2O.
Solution: moles Ca3N2 = 10.0 g Ca3N2 x = 0.06743 mole
1 mole Ca3N2
148.3 g Ca3N2
moles Ca(OH)2 = 0.06743 mole Ca3N2 x = 0.2023 mole 1 mole Ca3N2
3 moles Ca(OH)2
mass Ca(OH)2 = 0.2023 mole x = 15.0 g
74.1 g Ca(OH)2
1 mole Ca(OH)2
(e) What mass of NH3 can be formed from the reaction of 23.5 g of H2O with an excess of Ca3N2?
(Ans. 7.40 g)
(f) What mass of H2O is needed to make 454 g of Ca(OH)2? (Ans. 221 g)
(g) What volume of H2O is needed to react completely with 45.0 g of Ca3N2? (Ans. 32.8 mL
Explanation: