Question

Litium nitride reacts with water to form ammonia and lithium hydroxide, according to the following balanced chemical equation:Li3N (s) + 3 H2O (l) → NH3 (g) + 3 LiOH (aq)If 4.87 g of lithium reacts with 5.80 g of water, how much ammonia is produced before the reaction stops?Select one:a.5.80 gb.9.38 gc.1.82 gd.4.87 gCan you show steps to figure this out.

Answer 1

Answer;

`C`

Explanation;

Here, we want to get the mass of ammonia produced

Firstly, we need to identify the limiting reactant

The limiting reactant is the reactant that produces less amount of the target product

Firstly, let us get the number of moles of each of the reactants that reacted

To get this, we divide the given mass of reactant by the molar mass of the reactant

We start with lithium Nitride

The molar mass of lithium nitride is 35 g/mol

Thus, the number of moles of lithium nitride that reacted will be:

`(4.87)/(35)\text{ = 0.139 mol}`

Since the number of moles in the equation of reaction for lithium nitride and ammonia is the same, the number of moles of ammonia produced will be also 0.139 mol

Now, let us get for water

The molar mass of water is 18 g/mol

Thus, the number of moles of water that reacted will be:

`(5.8)/(18)\text{ = 0.322 mol}`

From the equation of reaction:

3 moles of water produced 1 mole of ammonia

0.322 mol of water will produce:

`(0.322)/(3)\text{ = 0.1074 mol}`

Since water produces less number of moles, it is the limiting reactant

Finally, to get the mass of ammonia produced, we multiply the above by the molar mass of ammonia which is 17 g/mol

We have that as:

`0.1074\text{ }*\text{ 17 = 1.826 g}`