How many liters of hydrogen will it take to reduce 13.7 g of tin oxide (4)?

Let’s execute the solution:

According to the condition of the problem, we compose the process equation:
m = 13.7 g X l -?

SnO2 + 2H2 = Sn + 2H2O – OBP, hydrogen is released.

We make calculations:
M (SnO2) = 150.6 g / mol;

M (H2) = 2 g / mol;

Y (SnO2) = m / M = 13.7 / 150.6 = 0.09 mol.

Proportion:
0.09 mol (SnO2) – X mol (H2);

-1 mol -2 mol hence, X mol (H2) = 0.09 * 2/1 = 0.18 mol.

We find the volume of H2:
V (H2) = 0.18 * 22.4 = 4.03 L

Answer: you need hydrogen with a volume of 4.03 liters