Question

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

Answer 1

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}=(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
`(1)/(3)* 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.

Answer 2

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