Zn(s) + 2 HCl(aq) H2(g) + ZnCl2(aq)
When 25.0 g of Zn reacts, how many L of H2 gas are formed at STP?
1. 0.382 L
2. 0.0171 L
3. 22.4 L
4. 8.56 L
5. 4.28 L
Number 4
Here is a worked example. Just follow the steps. At the end you may convert moles H2 to volume by L = moles x 22.4.
http://www.jiskha.com/science/chemistry/stoichiometry.html
4. 8.56L of H2
To determine the volume of H2 gas formed at STP when 25.0 g of Zn reacts, we first need to calculate the number of moles of Zn using its molar mass (65.38 g/mol).
Number of moles of Zn = Mass of Zn / Molar mass of Zn
= 25.0 g / 65.38 g/mol
= 0.3822 mol (rounded to four decimal places)
According to the balanced chemical equation, the stoichiometric ratio between Zn and H2 is 1:1. This means that for every 1 mole of Zn, 1 mole of H2 is produced.
Therefore, the number of moles of H2 gas produced is also 0.3822 mol.
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L of volume.
Hence, the volume of H2 gas formed at STP when 25.0 g of Zn reacts is 0.3822 L.
Therefore, the correct answer is option 1. 0.382 L.