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Answer 1 · 2016-10-08T19:04:06

$"0.118 L"$

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 $rarr$ aluminium sulfate + hydrogen)

$2"Al" + 3"H"_2"SO"_4 rarr "Al"_2("SO"_4)_3 + 3"H"_2$

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

moles of $"Al"$ = $("mass")/"A"_"r"= 12.7/27.0 = 0.470$

molar ratio of $"Al":"H"_2"SO"_4$ = $2:3$ ; $therefore$
moles of $"H"_2"SO"_4 = 0.470*(3/2) = 0.705$

$3.$ Solve for the unknown:

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

$6.00=0.705/"L"$

$"6.00 L"=0.705$

$"L"=0.705/6.00 = 0.118$