# 1. Lithium are nitrogen react in a combination reaction to produce lithium nitride: 6Li(s) +...

## Question:

1. Lithium are nitrogen react in a combination reaction to produce lithium nitride:

6Li(s) + N_2(g) rightarrow 2Li_3N(s)

In a particular experiment, 1 00 g samples of each reagent are reacted. The theoretical yield of Iithium nitride is ........ g.

a) 5.0

b) 0.84

c) 1.01

d) 1.67

e) 2.50

2. Balance the following equation.

............C_10H_12 + ............O_2 ----->........H_2O + ........CO_2

3. Balance the following equation.

........C_9H_20 + ...........O_2 ----> .......H_2O + .......CO_2

## Balancing chemical equations

A chemical equation represents a chemical reaction in form of symbols and formulas. The reactants are written on the left-hand side while the products are written on the right-hand side. In order to balance a chemical equation, you need to count the atoms of each element in the reactants and the products and ensure the number of elements in the reactant's side is equal to the number of elements in the product's side.

1.

{eq}6Li_{(s)} + N_2{(g)} \rightarrow 2Li_3N_{(s)} {/eq}

In the above reaction,

6 moles of Li reacts with 1 mole of {eq}N_2 {/eq} produces 2 moles of {eq}Li_3N {/eq}.

The molar mass of Li = {eq}7 g/mol {/eq}

7 g of Li = 1 mol

1 g of Li ={eq}\frac { 1}{7 mol} = 0.143 mol {/eq}

Molar mass of {eq}N_2 = 28 g/mol {/eq}

28 g of N_2 = 1 mol

1 g of {eq}N_2 = \frac {1}{28} = 0.036 mol {/eq}

Since {eq}Li {/eq} is used as excess, {eq}N_2 {/eq} is the limiting reagent.

1 mole of {eq}N_2 {/eq} produces 2 moles of {eq}Li_3N {/eq}.

Molar mass of {eq}Li_3N {/eq} = {eq}34.83 \:g/mol {/eq}

1 mole of {eq}Li_3N {/eq} = 34.83 g

0.072 moles of {eq}Li_3N {/eq} = {eq}0.072 \times 34.83 \:g {/eq} = 2.51 g

Theoritical Yield ={eq}2.51 g {/eq}

2.

{eq}C_{10}H_{12} + O_2 \rightarrow CO_2 + H_2O {/eq}

First balance the C by multiplying {eq}CO_2 {/eq} by 10

{eq}C_{10}H_{12} + O_2 \rightarrow 10CO_2 + H_2O {/eq}

balance the H by multiplying {eq}H_2O {/eq} by 6

{eq}C_{10}H_{12} + O_2 \rightarrow 10 CO_2 + 6 H_2O {/eq}

balance the O by multiplying {eq}O_2 {/eq} by 13

{eq}C_{10}H12 + 13 O_2 \rightarrow 10 CO_2 + 6 H_2O {/eq}

3.

{eq}C_9H_{20} + O_2 \rightarrow CO_2 + H_2O {/eq}

The Balanced equation is:

{eq}C_9H_{20} + 14 O_2 \rightarrow 9 CO_2 + 10 H_2O {/eq}