# Lithium metal will react readily with sulfuric acid. Which of the following is correct statement about the quantities of each reactant needed to produce the maximum amount of products without any reactant remaining?

## Answer is: a) The mass ratio of lithium to sulfuric acid is 1:7 Why?

Aug 2, 2017

Here's why that is the case.

#### Explanation:

Lithium metal will react with sulfuric acid to produce lithium sulfate and hydrogen gas as described by the balanced chemical equation

$2 {\text{Li"_ ((s)) + "H"_ 2"SO"_ (4(aq)) -> "Li"_ 2"SO"_ (4(aq)) + "H}}_{2 \left(g\right)} \uparrow$

Notice that the reaction consumes $1$ mole of sulfuric acid for every $2$ moles of lithium that take part in the reaction.

This means that the two reactants take part in the reaction in a $2 : 1$ mole ratio. As you know, you can convert this mole ratio to a gram ratio by using the molar masses of the two reactants.

In this case, you will have

• ${M}_{\text{M Li" = "6.941 g mol}}^{- 1}$
• M_ ("M H"_2"SO"_4) = "98.079 g mol"^(-1)

You can thus say that the $2 : 1$ mole ratio will be equivalent to

"2 moles Li"/("1 mole H"_2"SO"_4) = (2 color(red)(cancel(color(black)("moles Li"))) * "6.941 g"/(1color(red)(cancel(color(black)("mole Li")))))/(1 color(red)(cancel(color(black)("mole H"_2"SO"_4))) * "98.079 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4))))) = "13.882 g Li"/("98.079 g H"_2"SO"_4)

This, of course, is equal to

"13.882 g Li"/("98.079 g H"_2"SO"_4) = "1 g Li"/("7.065 g H"_2"SO"_4) ~~ "1 g Li"/("7 g H"_2"SO"_4)

Therefore, you can say that the two reactants take part in this reaction in a $2 : 1$ mole ratio and in a $1 : 7$ gram ratio, or mass ratio, which means that for every gram of lithium that takes part in the reaction, the reaction consumes $\text{7 g}$ of sulfuric acid.