what volume of hydrogen at STP is released when 8.25g Mg reacts?

Question

Mg metal reacts with HCl to produce hydrogen gas:

Mg(s) + 2Hcl(ag)-----> MgCl2(ag) + H2 (g)

what volume of hydrogen at STP is released when 8.25g Mg reacts?

Answer 1

Mg Mw = 24.3 g/mole

8.25 g = 8.25 g * mole/24.3g = 0.3395 mole

1:1 releationship Mg:H2 production

so would expect 0.3395 mole H2 MW = 2 g/mole
or

V = nRT/P
V= 0.3395mole(0.08205 l-atm/mole-K)273K/1atm
= 7.6 liters

Answer 2

What do we know?

1) 8.25 g Mg reacts with HCl to yield MgCl2 and H2.

2) MW of Mg = 24.3 g per mole (from Periodic Table)

3) Mg and H2 have a 1:1 mole relationship

4) Recall that the volume of an ideal gas (regardless of composition) at STP is 22.4 L per mole.

What are we looking for?: volume of H2 released at STP

8.25 g Mg x (1 mole Mg / 24.3 g Mg) x (1 mole H2 / 1 mole of Mg) x (22.4 L H2 / 1 mole of H2) = 7.60 L H2

Enjoy!