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STUCK ON THIS!!
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

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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 [tex]Li_3N[/tex] will react to produce 120 grams of LiOH.

Explanation: Moles can be calculated by using the formula:

[tex]Moles=\frac{\text{given mass}}{\text{molar mass}}[/tex]    ...(1)

Molar mass of LiOh = 23.95 g/mol

Given mass of LiOH = 120g

[tex]\text{Moles of LiOH}=\frac{120g}{23.95g/mol}=5.01moles[/tex]

For a given chemical reaction:

[tex]Li_3N(s)+3H_2O\rightarrow NH_3(g)+3LiOH(l)[/tex]

By Stoichiometry,

3 moles of LiOH is produced by 1 mole of [tex]Li_3N[/tex]

5.01 moles of LiOH will be produced by [tex]\frac{1}{3}\times 5.01[/tex] =1.67 moles of [tex]Li_3N[/tex]

Mass of [tex]Li_3N[/tex] will be calculated using the equation 1:

[tex]1.67mol=\frac{\text{Given mass}}{34.83g/mol}[/tex]

Mass of [tex]Li_3N[/tex] = 58.166 grams.

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