When a mixture of iron and copper was treated with a solution of sulfuric acid, 1.12 liters of colorless gas were released

When a mixture of iron and copper was treated with a solution of sulfuric acid, 1.12 liters of colorless gas were released, and when the same sample was treated at room temperature with concentrated nitric acid, 22.4 liters of brown gas were formed. Initial mixture weight

Let’s write down given:

Fe + Cu mixture

+ H2SO4 (solution)

V (gas) = ​​1.12 l

Fe + Cu mixture

+ HNO3 (conc.)

V (gas) = ​​22.4 l

m (mixture) -?

Decision:

1) Copper does not interact with sulfuric acid solution. When iron interacts with a sulfuric acid solution, a colorless gas is formed – hydrogen.

Let’s write down the reaction equation and calculate the mass of iron in the mixture:

Fe + H2SO4 = FeSO4 + H2

1 mol 1 mol

M (Fe) = 56 g / mol

Vm = 22.4 l / mol

56 g (Fe) forms 22.4 L (H2)

X g (Fe) forms 1.12 L (H2)

X = 56 g * 1.12 l / 22.4 l

X = 2.8 g (Fe)

2) Iron does not interact with concentrated nitric acid. When copper interacts with concentrated nitric acid, a brown gas is formed – nitric oxide (IV).

Let’s write down the reaction equation and calculate the mass of copper in the mixture:

Cu + 4 HNO3 = Cu (NO3) 2 + 2 NO2 + 2 H2O

1 mol 2 mol

M (Cu) = 64 g / mol

64 g (Cu) forms 2 * 22.4 L (NO2)

X g (Cu) forms 22.4 L (NO2)

X = 64 g * 22.4 l / 44.8 l

X = 32g (Cu)

3) Find the mass of the mixture:

m (mixture) = 2.8 g + 32 g = 34.8 g

Answer: 34.8 g