Question

Lithium and nitrogen react in a combination reaction to produce lithium nitride:

             6Li(s) + N2(g) → 2Li3N(s)

In a particular experiment, 3.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

18) Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.

                        2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

A) 20.00%

B) 25.00%

C) 50.00%

D) 12.50%

Answer 1

1)

Molar mass of Li = 6.968 g/mol

mass of Li = 3.5 g

we have below equation to be used:

number of mol of Li,

n = mass of Li/molar mass of Li

=(3.5 g)/(6.968 g/mol)

= 0.5023 mol

Molar mass of N2 = 28.02 g/mol

mass of N2 = 3.5 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(3.5 g)/(28.02 g/mol)

= 0.1249 mol

we have the Balanced chemical equation as:

6 Li + N2 ---> 2 Li3N

6 mol of Li reacts with 1 mol of N2

for 0.5023 mol of Li, 0.0837 mol of N2 is required

But we have 0.1249 mol of N2

so, Li is limiting reagent

we will use Li in further calculation

Molar mass of Li3N = 3*MM(Li) + 1*MM(N)

= 3*6.968 + 1*14.01

= 34.914 g/mol

From balanced chemical reaction, we see that

when 6 mol of Li reacts, 2 mol of Li3N is formed

mol of Li3N formed = (2/6)* moles of Li

= (2/6)*0.5023

= 0.1674 mol

we have below equation to be used:

mass of Li3N = number of mol * molar mass

= 0.1674*34.91

= 5.85 g

Answer: 5.85 g

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