Question

What volume of hydrogen at STP is released when 8.35 g Mg reacts? (Assume that the amount of HCl is enough for the reaction.)

Answer 1

Write the reaction:

$$\text{Mg} + 2 \text{HCl} → \text{MgCl}_2+\text{H}_2\uparrow$$

As we see, one atom of magnesium is required to produce one molecule of hydrogen (gas). Therefore, first, we need find the amount of substance of Mg:

$$n(\text{Mg})=\frac{m}{\Mu}=\frac{8.35}{24}=0.35\text{ mol}.$$

Hence, 0.35 mol of hydrogen will be produces (the number of hydrogen molecules is the same as the number of magnesium atoms).

So, volume of any gas is the amount of substance times molar volume:

$$V(\text{H}_2)=nV_m=0.35\cdot22.4=7.8\text{ L}.$$