Question

What volume (in L) of hydrogen gas is produced at STP when 45.1 g of Zn metal reacted completely with an excess of H2SO4?

Answer 1

The balanced reaction for this reaction is:

`Zn_((s))+H_2SO_(4(aq))\rightarrow ZnSO_(4(aq))+H_(2(g))`

Step 1: We will convert the mass of the Zn metal to moles:

`\begin{gathered} moles=\frac{mass}{molar\text{ }mass} \\ \\ moles=(45.1g)/(65.38gmol^(-1)) \\ \\ moles=0.689813\text{ }moles\text{ }Zn \end{gathered}`

Step 2: We will determine the moles of hydrogen gas that is produced from 0.689813 moles of Zn:

`\begin{gathered} mole\text{ }ratio=1\text{ }Zn:1\text{ }H_2 \\ Therefore,0.689813\text{ }moles\text{ }Zn=0.689813\text{ }moles\text{ }H_2 \end{gathered}`

Step 3: In order to determine the volume of gas we use the ideal gas law: