0.442 L of hydrogen sulfide was passed through 150 g of 5% lead solution (2).

| September 6, 2021 | education

0.442 L of hydrogen sulfide was passed through 150 g of 5% lead solution (2). Calculate the mass fraction of nitric acid in the resulting solution.

Given:
m solution (Pb (NO3) 2) = 150 g
ω (Pb (NO3) 2) = 5%
V (H2S) = 0.442 l

Find:
ω (HNO3) -?

Solution:
1) Pb (NO3) 2 + H2S => PbS + 2HNO3;
2) m (Pb (NO3) 2) = ω (Pb (NO3) 2) * m solution (Pb (NO3) 2) / 100% = 5% * 150/100% = 7.5 g;
3) n (Pb (NO3) 2) = m (Pb (NO3) 2) / M (Pb (NO3) 2) = 7.5 / 331 = 0.0227 mol;
4) n (H2S) = V (H2S) / Vm = 0.442 / 22.4 = 0.0197 mol;
5) n (HNO3) = n (H2S) * 2 = 0.0197 * 2 = 0.0394 mol;
6) m (HNO3) = n (HNO3) * M (HNO3) = 0.0394 * 63 = 2.4822 g;
7) m (H2S) = n (H2S) * M (H2S) = 0.0197 * 34 = 0.6698 g;
8) m solution = m solution (Pb (NO3) 2) + m (H2S) = 150 + 0.6698 = 150.6698 g;
9) ω (HNO3) = m (HNO3) * 100% / m solution = 2.4822 * 100% / 150.6698 = 1.647%.

Answer: The mass fraction of HNO3 is 1.647%.



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