The volume of hydrogen required for the complete reduction of iron oxide (2) obtained by thermal

The volume of hydrogen required for the complete reduction of iron oxide (2) obtained by thermal decomposition of 36 g of iron nitrate (11).

To solve this problem, we write down given: m (Fe (NO3) 2) = 36 g.
Find: the volume of hydrogen.
Solution:
Let’s write down the first reaction equation.
Fe (NO3) 2 = Fe2O3 + NO2 + O2
Let’s arrange the coefficients:
4Fe (NO3) 2 = 2Fe2O3 + 8NO2 + O2
Let’s calculate the molar masses of iron (II) nitrate and iron (III) oxide.
M (Fe (NO3) 2) = 56 + (14 + 16 * 3) * 2 = 180 g / mol
M (Fe2O3) = 56 * 2 + 16 * 3 = 160 g / mol
Over the iron (II) nitrate we write 36 g, under the iron (II) nitrate – 720 g.
We write x g over iron (III) oxide, 320 g under iron (III) oxide.
Let’s compose and solve the proportion:
x = 36 * 320/720 = 16 g
Let’s write the second reaction equation:
Fe2O3 + H2 = H2O + Fe
Let’s arrange the coefficients:
Fe2O3 + 3H2 = 3H2O + 2Fe
We write 16 g over iron (III) oxide, 160 g under iron (III) oxide.
Over hydrogen we write u l, under hydrogen – 67.2 l
Let’s compose and solve the proportion:
y = 16 * 67.2 / 160 = 6.72 liters.
Answer: 6.72 L