Question

40% of nitric acid reacts with 4.11g of lead carbonate of 65% purity.Calculate:
1. weight of pure lead carbonate in 4.11g.
2. weight of pure nitric acid required to dissolve lead carbonate.
3. weight of 40% nitric acid required. ​

Answer 1

Answer:

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Answer 2

1. Weight of pure lead carbonate is 2.6715 g.
2. Weight of pure nitric acid required is 1.26 g.
3. Weight of 40% nitric acid required is 3.15 g.

• Weight of pure lead carbonate = 65×4.11/100 g.
• Moles of lead carbonate = 2.6715/267.15 = 0.01 moles                                             ( Molar mass=267.15 g/mole)
• Moles of nitric acid required = 2×0.01 = 0.02 moles
• weight of pure nitric acid required = 0.02×63=1.26 g.
• Weight of 40% nitric acid required = 1.26×100/40=3.15 g.