Question

How many grams of LigN can be formed from 1.75 moles of Li? Assume an excess of nitrogen.6 Li(s) + N2(g)-› 2 Li3N(s)O 20.3 g LizN© 93.5 g LigNO 4.56 g LizNO 10.3 g Li3N

Answer 1

Answer: first option (or letter A): 20.3 g LizN

Step-by-step explanation:

The question requires us to calculate the mass of Li3N produced when 1.75 moles of Li and excess of N2 are used in the following chemical reaction:

`6Li_((s))+N_(2(g))\rightarrow2Li_3N_((s))`

To solve this problem, we must consider the stoichiometry of the reaction to calculate the amount of moles of Li3N produced and then use the molar mass of Li3N (34.83 g/mol) to convert the number of moles in mass of Li3N.

According to the chemical equation, 2 moles of Li3N are produced when 6 moles of Li are reacting. Thus, we can write:

6 mol Li ---------------------- 2 mol Li3N

1.75 mol Li ------------------ x

Solving for x, we'll have:

`x=\frac{(1.75mol\text{ Li})*(2mol\text{ L}i_3N)}{(6mol\text{ Li})}=0.583mol\text{ L}i_3N`

Therefore, 0.583 moles of Li3N are obtained from 1.75 moles of Li.

Now, considering the molar mass of Li3N (34.83 g/mol):

1 mol Li3N --------------- 34.83g Li3N

0.583 mol Li3N ------- y

Solving for y, we'll have that 0.583 moles of Li3N correspond to 20.3 g of this compound.

The best option to answer this question is the first one (or letter A): 20.3 g LizN.