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.
Answers
Answered by
0
Answer:
you can search it on Google
Explanation:
follow me n I will follow back u
Answered by
1
Quantity of Nitric acid = 40% (Given)
Quantity of lead carbonate = 65% (Given)
We know that,
1. weight of pure lead carbonate in 4.11g
= 65 × 4.11/100
= 2.6715
Moles of lead carbonate
= 2.6715/267.15
= 0.01 moles
2. Moles of nitric acid required
= 2 ×0.01
= 0.02 moles
Weight of pure nitric acid =
= 0.02 × 63
= 1.26 g.
3. Weight of 40% nitric acid required
= 1.26 × 100/40
= 3.15 g.
Similar questions