Question

How many grams of hydrogen gas can be produced from the following reaction when 65 grams of zinc and 65 grams of $\text{HCl}$ are allowed to react.
A. 1.0
B. 1.8
C. 3.6
D. 7.0
E. 58

Answer 1

Hint: The mass of zinc and hydrochloric acid is given in the question and we can calculate the molar mass easily so by using this information we can calculate the number of moles of zinc and hydrochloric acid and hydrogen.

Complete step by step answer:
- In the given question, we have to find the mass of hydrogen gas which is produced by the zinc and hydrochloric acid.
- It is given in the question that the mass of zinc and hydrochloric acid is 65g.
- Now, the reaction of zinc with hydrochloric acid will be:
$\text{Zn + 2HCl }\to \text{ ZnC}{{\text{l}}_{2}}\text{ + }{{\text{H}}_{2}}$

- We can see that only one mole of zinc reacts with the two moles of hydrochloric acid and they produce one mole of hydrogen gas.
- Now, we will calculate the molar mass of hydrochloric acid that is:
$\text{HCl = 1 }\times \text{ 35}\text{.5 = 36}\text{.5 g/mol}$
- And we know that the atomic mass of zinc is 65.4 g/mol.

- So, now by applying the formula of no. of moles we will get:
$\text{No}\text{. of moles = }\dfrac{\text{Mass}}{\text{Molar mass}}$
$\text{No}\text{. of moles (Zn) = }\dfrac{65}{65.4}\text{ = 0}\text{.994 moles}\text{.}$
$\text{No}\text{. of moles (HCl) = }\dfrac{65}{36.5}\text{ = 1}\text{.78 moles}\text{.}$
- So, we can say that according to the equation, 0.994 moles of zinc can react with $2\ \times \text{ 1}\text{.78 = 3}\text{.56 moles}$of $\text{HCl}$ but we have only 1.78 moles of $\text{HCl}$.
- So, here $\text{HCl}$ will act as a limiting reagent.
- So, 1.78 moles of $\text{HCl}$ will produce $\dfrac{1.78}{2}\ =0.89\text{ moles of }{{\text{H}}_{2}}$.

- So, the mass of hydrogen will be:
$\text{Mass = No}\text{. of moles }\times \text{ Molar mass}$
As we know the molar mass of hydrogen 2g ($2\ \times \text{1}$) and no. of moles are 0.89 moles, we will get: $\text{Mass = 0}\text{.89 }\times \text{ 2 = 1}\text{.78 }\sim \text{ 1}\text{.8g}$
So, the correct answer is “Option B”.

Note: In a chemical reaction, the limiting reactant is the amount of the reactant that is consumed completely in the reaction. Without the limiting reactant, the reaction is not able to proceed further and stops the reaction.