Question

In the equation:
4NH3 + 502 → 4NO + 6H2O
-
Find the volume of oxygen that reacts with 1.12 L of NH3.
0 0.896 L
O 1.4L
0 1.12
0 561

Answer 1

Question:-

Find the volume of oxygen that reacts with 1.12 L of NH3.

Equation:-

$4NH {\tiny{3}} + 5 0{ \tiny{2 }} \: \: -> \: 4 NO + 6H{ \tiny{2}}O$

Options :-

0 0.896 L

O 1.4L

0 1.12

0 561

Answer:- 1.4L

Given:-

1. 4 moles of NH3 reacts with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O .
2. 1.12 L is the amount of NO which have to react with O2

To find :- amount of O2 with will react with 1.12 L of NH3

Explanation:-

Formulas:-

1. Conservation of mass
2. $moles = \frac{volume \: of \: gas \: in \: lts}{22.4lts}$

According to the reaction:-

$4NH {\tiny{3}} + 5 0{ \tiny{2 }} \: \: -> \: 4 NO + 6H{ \tiny{2}}O$

4 moles of NH3 reacts with 5 moles of O2

so:-

1 mole of NH3 reacts with 5/4 moles of O2

Number of moles given of NH3 :-$\frac{1.12}{22.4} moles$

For given moles :-

$\frac{1.12}{22.4} mole \: of \: NH { \tiny3 }\: reacts \: with \: \frac{5}{4} \times \frac{1.12}{22.4} moles \: of \: O2$

So numbers of moles of O2 required:-

$\frac{5 \times 112}{100 \times 4 \times 22.4} \\ = \frac{5 \times \cancel{112} {}^{ \: \: 28} }{100 \times \cancel{ 4} \times 22.4} \\ = \frac{ \cancel5 \times \cancel{28} {}^{14} }{{\cancel{100} {}^{10} }\times 22.4} = \frac{1.4}{22.4} moles$

To convert it into volume of gas in lts just multiple it with 22.4 lts :-

$\frac{1.4}{22.4} \times 22.4 \: \: lts \\ \frac{1.4}{ \cancel{22.4}} \times \cancel{22.4} \: \: lts \\ 1.4lts \: is \: your \: answer$

Option :- Second :- 1.4 L