Question

What volume of hydrogen (in L) is produced

from the complete reaction of 56.49 g of
magnesium metal at STP?
(Mg = 24.30 g/mol)
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Hint: 1 mole of gas at STP occupies 22.4 L

Answer 1

52.07 L H₂

Step-by-step explanation:

Before you can find the volume of H₂, you need to find the moles of H₂. To do this, you need to (1) convert grams Mg to moles (via the atomic mass) and then (2) convert moles Mg to moles H₂ (via the mole-to-mole ratio from equation coefficients).

Atomic Mass (Mg): 24.30 g/mol

1 Mg(s) + 2 HCl(aq) ---> MgCl₂(aq) + 1 H₂(g)
^ ^

56.49 g Mg 1 mole 1 mole H₂
------------------- x ----------------- x -------------------- = 2.32 moles H₂
24.30 g 1 mole Mg

Now that you know the moles of H₂, you need to determine the volume at STP. To do this, you need to set up a proportion comparing the mole value versus the volume. Then, you can cross-multiply to solve for the unknown volume. The final answer should have 4 sig figs to match the given values.

1 mole 2.32 moles
-------------- = -------------------- <----- Set up proportion
22.4 L ? L

(1 mole) x ? L = 52.07 <----- Cross-multiply

? L = 52.07 <----- Divide both sides by 1 mole