When 25.0g of Zn reacts with Hcl, how many L of H2 gas are formed at STP?

Answer 1

To calculate the volume of H2 gas formed at STP, we must first use the balanced chemical reaction:

Zn + 2HCl → ZnCl2 + H2

Next, we need to determine the moles of Zn and HCl reacting:

25.0 g Zn * (1 mol Zn / 65.38 g Zn) = 0.382 mol Zn

Now we can use stoichiometry to find the moles of H2 gas formed:

0.382 mol Zn * (1 mol H2 / 1 mol Zn) = 0.382 mol H2

At STP (standard temperature and pressure), 1 mole of an ideal gas occupies a volume of 22.4 L. So, we can calculate the volume (L) of H2 gas:

0.382 mol H2 * (22.4 L H2 / 1 mol H2) = 8.56 L H2

So, 8.56 L of H2 gas are formed at STP when 25.0g of Zn reacts with HCl.