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consider the following reaction

Li3N (s) + 3H2o (l) --> NH3 (g) + 3LiOh (l)

if you need to make 120 g LiOh, how many grams of Li3N must you react with excess water


-700g

-117g

-350g

-58g

Verified Answer :

MissPhiladelphia

The balanced chemical reaction is:

Li3N (s) + 3H2o (l) --> NH3 (g) + 3LiOh (l)

We are given the amount of LiOH produced from the reaction. This will be the starting point of the calculations. We do as follows:

120 g LiOH ( 1 mol / 23.95 g ) ( 1 mol Li3N / 3 mol LiOH ) ( 34.83 g / 1 mol ) = 58.2 g

Answer: 58 grams of \(Li_3N\) will react to produce 120 grams of LiOH.

Explanation: Moles can be calculated by using the formula:

\(Moles=\frac{\text{given mass}}{\text{molar mass}}\)    ...(1)

Molar mass of LiOh = 23.95 g/mol

Given mass of LiOH = 120g

\(\text{Moles of LiOH}=\frac{120g}{23.95g/mol}=5.01moles\)

For a given chemical reaction:

\(Li_3N(s)+3H_2O\rightarrow NH_3(g)+3LiOH(l)\)

By Stoichiometry,

3 moles of LiOH is produced by 1 mole of \(Li_3N\)

5.01 moles of LiOH will be produced by \(\frac{1}{3}\times 5.01\) =1.67 moles of \(Li_3N\)

Mass of \(Li_3N\) will be calculated using the equation 1:

\(1.67mol=\frac{\text{Given mass}}{34.83g/mol}\)

Mass of \(Li_3N\) = 58.166 grams.

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