# Question #c67b3

Feb 13, 2016

6.3g

#### Explanation:

$M {g}_{\left(s\right)} + 2 H C {l}_{\left(a q\right)} \rightarrow M g C {l}_{2 \left(a q\right)} + {H}_{2 \left(g\right)}$

This tells us that 1 mole of magnesium produces 1 mole of hydrogen gas.

So we can say that 3.15 moles of magnesium must produce 3.15 moles of hydrogen gas.

Hydrogen gas has the molecular formula ${H}_{2}$.

The ${A}_{r}$ of hydrogen = 1.

So the ${M}_{r}$ of hydrogen gas ${H}_{2}$ = 1 + 1 =2

So the mass of 1 mole of hydrogen gas = 2g

So the mass of 3.15 moles of hydrogen gas = 3.15 x 2 = 6.3g.

Please note that you did not fully balance the equation.
Since you only have one Cl on the left hand side yet you have two Cl on the right, you need one more on the left. If you add one more, you now have 2 Cl. So you add the 2 infront of HCl, like this:
$M g + 2 H C l \implies M g C {l}_{2} + {H}_{2}$

Now double check to see if they are fully balanced. Count each atoms on both sides to see if they have the same number, like this:
1 Mg on the left, 1 Mg on the right - balanced
2 H on the left, 2 H on the right - balanced
2 Cl on the left, 2 Cl on the right - balanced
Now that they're all balanced, your equation is now fully balanced. In future, check that you have included all atoms when counting them, don't miss out any.

So, back to Stoichiometry.
Assume that: (n) - number of moles, (m) - mass of substance, and (M) - molar mass.
Note that: $n = m \div M$

Since 3.15 moles of Mg is provided for you, you need to find out how many moles of ${H}_{2}$ will be given. Looking back at the equation, the mole ratio between Mg : H is 1 : 1.
If 1 mole of Mg gives you 1 mole of H, then 3.15 moles of Mg should give you: [$3.15 \div 1 \times 1$] = 3.15 moles of H.

Now you've found the number of moles for H, you now need to find the molar mass (M) of H. If you look into your periodic table, the molar mass (or molecular weight) of H is 1.0 g/mol.

Since there are 2 H present, the molar mass (M) of ${H}_{2}$ is 1.0 g/mol $\times$ 2 = 2.0 g/mol

The mass (m) of ${H}_{2}$ is:
$m = M \times n$
[$m = 2.0 \times 3.15$] = 6.30 grams.

Therefore, 6.30 grams of ${H}_{2}$ is produced by the reaction of 3.15 moles of Mg.