# Question 6f712

Oct 8, 2016

$\text{0.118 L}$

#### Explanation:

To work out how many litres of sulfuric acid we need, we have to first work out the moles of sulfuric acid we have reacting in our equation.

$1.$ Write a balanced equation:

(aluminium $+$ sulfuric acid $\rightarrow$ aluminium sulfate + hydrogen)

$2 {\text{Al" + 3"H"_2"SO"_4 rarr "Al"_2("SO"_4)_3 + 3"H}}_{2}$

$2.$ Work out the moles of one reagent and use the molar ratio to work out the moles of the other:

moles of $\text{Al}$ = ("mass")/"A"_"r"= 12.7/27.0 = 0.470#

molar ratio of ${\text{Al":"H"_2"SO}}_{4}$ = $2 : 3$; $\therefore$
moles of ${\text{H"_2"SO}}_{4} = 0.470 \cdot \left(\frac{3}{2}\right) = 0.705$

$3.$ Solve for the unknown:

We know that $\text{M" = "m"/"L}$, so we can now solve for $\text{L}$:

$6.00 = \frac{0.705}{\text{L}}$

$\text{6.00 L} = 0.705$

$\text{L} = \frac{0.705}{6.00} = 0.118$