How much hydrogen is required to obtain chromium from 1 kg of ore containing 76% chromium oxide.

univerkov | February 13, 2021 | education

Data: mр – mass of taken ore (mр = 1 kg); ωCr2O3 is the content of chromium III oxide in the taken ore (ωCr2O3 = 76% = 0.76).

Const: MCr2O3 is the molar mass of chromium III oxide (MCr2O3 = 152 g / mol); Vm – molar volume (for n.o. Vm = 22.4 l / mol).

1) Find out the mass of the reacted chromium III oxide: mCr2O3 = mр * ωCr2O3 = 1 * 0.76 = 760 g.

2) The considered reaction: 2Cr (chromium) + 3H2O (water) = Cr2O3 (chromium oxide) + 3H2 (hydrogen).

3) Amount of chromium oxide III substance: νCr2O3 = mCr2O3 / MCr2O3 = 760/152 = 5 mol.

4) Amount of hydrogen substance: νH2 / νCr2O3 = 3/1 and νH2 = 3 * νCr2O3 = 3 * 5 = 15 mol.

5) The volume of hydrogen: VH2 = Vm * νH2 = 22.4 * 15 = 336 liters.

Answer: You will need 336 liters of hydrogen.

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