Calculate the volume of hydrogen that will be released as a result of the interaction of 45.5 g

Calculate the volume of hydrogen that will be released as a result of the interaction of 45.5 g of zinc with a solution of H2TeO4.

1. Let’s compose the equation:
m = 45.5 g. V l. -?
Zn + H2TeO4 = H2 + Zn (TeO4) – substitution reaction, hydrogen is released;
2. Calculation:
M (Zn) = 65.3 g / mol;
M (H2) = 2 g / mol.
3. Determine the number of moles of the starting substance, if the mass is known:
Y (Zn) = m / M = 45.5 / 65.3 = 0.7 mol;
Y (H2) = 0.7 mol since the amount of these substances according to the equation is 1 mol.
4. Find the volume of the product:
V (H2) = 0.7 * 22.4 = 15.68 liters.
Answer: during the reaction, a volume of 15.68 liters of hydrogen was obtained.