Question

What volume of Hz is produced at 299 K and 1.07 atm when 55.8 g of Zn metal react with excess

HCl?
Zn(s) + 2 HCI(aq) —› ZnCh(aq) + H2(g)

Answer 1

Moles of zn

• 55.8/65
• 0.86mol

1 mol produces 1 mol H2

• Moles of H_2=n=0.86mol
• P=1.07atm
• T=299K

So

Ideal gas equation

• PV=nRT
• V=nRT/P
• V=0.86(8.314)(299)/1.07
• V=1.998L

Answer 2

2.29 L

Step-by-step explanation:

In this question we have to start with the chemical reaction:

The reaction is already balanced. So, if we have an excess of HCl the compound that would limit the production of would be Zn. So, we have to follow a few steps:

1) Convert from grams to moles (Using the atomic mass of Zn 65.38 g/mol).

2) Convert from moles of Zn to moles of (Using the molar mass 1 mol = 1 mol Zn).

3) Convert from mol of to volume (Using the ideal gas equation PV=nRT).

We have to remember that R = 0.082