$$ \ce{H2 + F2 -> 2HF} $$

In the reaction represented above, what mass of \(\ce{HF}\) is produced by the reaction of \(3.0 \times 10^{23}\) molecules of \(\ce{H2}\) with excess \(\ce{F2}\) ? (Assume the reaction goes to completion.)

Summary Submit

Since one reactant is given in excess, we can simply convert the given quantity using stoichiometry. We use Avogadro's number to convert molecules into moles.

$$ 3.0 \times 10^{23} \text{ molecules }\ce{H2} \cdot \frac{\pu{1 mol }\ce{H2}}{6.02 \times 10^{23} \text{ molecules }\ce{H2}} \cdot \frac{\pu{2 mol }\ce{HF}}{\pu{1 mol }\ce{H2}} \cdot \frac{\pu{20 g }\ce{HF}}{\pu{1 mol }\ce{HF}}$$ $$ \approx \boxed{\pu{20 g}} $$