Calculate the volume of hydrogen required to reduce 1 ton of ore containing 92% tungsten oxide (VI).

Calculate the volume of hydrogen required to reduce 1 ton of ore containing 92% tungsten oxide (VI). What is the mass of tungsten, which yield is 78%?

1) According to the condition of the problem, hydrogen, tungsten (VI) oxide and tungsten are given. Using the periodic system of DI Mendeleev, calculate the molar masses of these substances. They will be numerically equal to molecular weights:
Mr (H2) = Ar (H) * 2 = 2 g / mol;
Mr (WO3) = Ar (W) + Ar (O) * 3 = 184 + 16 * 3 = 232 g / mol;
Mr (W) = Ar (W) = 184 g / mol;

2) In the task, the unit of mass is indicated in the ton. Convert to grams:
m (ore) = 1 t = 1,000,000 g;

3) Write down the equation of the ongoing chemical reaction:
WO3 + 3H2 => W + 3H2O;

4) Calculate the mass of the substance WO3:
m (WO3) = m (ores) * ω (WO3) / 100% = 1,000,000 * 92% / 100% = 920000 g;

5) Find the amount of substance WO3:
n (WO3) = m (WO3) / Mr (WO3) = 920000/232 = 3965.5 mol;

Finding the volume of hydrogen
6) Find the amount of substance H2 (taking into account the reaction equation):
n (H2) = 3 * n (WO3) = 11896.5 mol;

7) Calculate the required volume of H2:
V (H2) = n (H2) * Vm = 11896.5 * 22.4 = 266481.6 l;

Determination of the mass of tungsten
8) Determine the amount of substance W (taking into account the reaction equation):
n (W) = n (WO3) = 3965.5 mol;

9) Find the theoretical mass of matter W:
m theor. (W) = n (W) * Mr (W) = 3965.5 * 184 = 729652 g;

10) Calculate the practical mass of the substance W:
m pract. (W) = m theore. (W) * ω out. (W) / 100% = 729652 * 78% / 100% = 569129 g.

Answer: The volume of H2 is 266481.6 liters; practical weight W – 569129 g.