Question

If sufficient acid is used to react completely
with 21.0 grams of Mg

Mg(s) + 2 HCl(aq) + MgCl2(aq) + H2(g)

what volume of hydrogen at STP would be
produced?

1. 9.68 liters
2. 10.60 liters
3. 4.84 liters
4. 19.37 liters
5. 22.40 liters

Answer 1

Solution,

Mass of Mg - 21gm

molar mass of Mg - 24gm

moles = Given mass/ Molar mass

moles = 21/24 = 0.875

1 mole of Mg produce 1 mole of H2 gas

so 0.875 mole of Mg will produce 0.875 moles of H2 gas

One mole of H2 gas = 22.4 litre

0.875 mole of H2 gas = 0.875×22.4

0.875 mole of H2 gas = 19.60 litre

So the nearest ANSWER is Option four 19.37 litre.

Answer 2

STP is the standard temperature and the pressure of the gases of the system. The volume of the hydrogen that will be produced as STP is 19.37 litres.

What is volume?

Volume is the amount of the substance or the solution occupied in an area or object and is estimated in L or mL. It is a scalar quantity.

Given,

Mass of magnesium = 21 gm

Molar mass of magnesium = 24 gm/mol

Moles of magnesium can be calculated as,

$\begin{aligned}\rm moles &= \rm \dfrac{mass}{molar\;mass}\\\\&= \dfrac{21}{24}\\\\&= 0.875 \;\rm mol\end{aligned}$

If 1-mole magnesium = 1-mole hydrogen gas

Then 0.875-mole magnesium = 0.875 moles of hydrogen gas

If 1 mole of hydrogen gas = 22.4 L

Then, 0.875 moles of hydrogen gas = x L

Solving for x:

$\begin{aligned}\rm x &= 0.875 \times 22.4\\\\&= 19.60 \;\rm litre\end {aligned}$

Therefore, the volume of hydrogen at STP would be option 4. 19.37 litres.

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