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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

1 Answer

14 votes

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 = (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 :-
(1.12)/(22.4) moles

For given moles :-


(1.12)/(22.4) mole \: of \: NH { \tiny3 }\: reacts \: with \: (5)/(4) * (1.12)/(22.4) moles \: of \: O2

So numbers of moles of O2 required:-


(5 * 112)/(100 * 4 * 22.4) \\ = \frac{5 * \cancel{112} {}^( \: \: 28) }{100 * \cancel{ 4} * 22.4} \\ = \frac{ \cancel5 * \cancel{28} {}^(14) }{{\cancel{100} {}^(10) }* 22.4} = (1.4)/(22.4) moles

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


(1.4)/(22.4) * 22.4 \: \: lts \\ \frac{1.4}{ \cancel{22.4}} * \cancel{22.4} \: \: lts \\ 1.4lts \: is \: your \: answer

Option :- Second :- 1.4 L

User Karun
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