The mixture of ammonia and carbon monoxide was divided into two equal parts. When passing the first

| August 14, 2021 | education

The mixture of ammonia and carbon monoxide was divided into two equal parts. When passing the first part through a tube with incandescent copper (II) oxide, the mass of the tube decreased by 8.8 g, the second part can completely neutralize 147 g of 10% sulfuric acid. fraction of gases in the initial mixture

Let’s compose the reaction equalization:
CO + CuO = Cu + CO2. Reaction type: Double replacement
2NH3 + 3CuO = Na2 + 3Cu + 3H2O
2NH3 + H2SO4 = (NH4) 2SO4
The amount of ammonia substance is equal to:
V (H2SO4) = (147 * 0.1) 98 = 0.15 mol
V (NH3) = 2v (H2SO4) = 0.3 mol
The amount of carbon monoxide substance is equal to:
8.8 = m (CuO) -m (Cu) = m (O)
V (O) = 8.8 / 16 = 0.55 mol = V (CuO)
V (1 CuO) = 3 / 2V (NH3) = 0.45 mol
V (2CuO) = 0.55-0.45 = 0.1 mol
v (CO) = V2 (CuO) = 0.1 mol
Volume fractions of gases in the mixture
Omega (CO) = 0.1 / (0.1 + 0.3) * 100% = 25%
Omega (NH3) = 100% -25% = 75%
Answer: the volume fractions of gases were 25% and 75%



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