Question

Can someone please answer this?

Answer 1

ANSWER :- 32.4 g H20

$\bf\red{ 4NH_{3} + 5O_{2} \rightarrow 4NO +6H_{2}O }$

$\bf\huge \underline{\underline{ Given :- }}$

• 51 g $\bf NH_{3}$

• 48 g $\bf O_{2}$

$\bf\huge\underline{\underline{ To \: Find:- }}$

• amount of $H_{2} O$ formed

$\bf\huge\underline{\underline{ Answer :- }}$

$\bf\large\underline{\underline{ Step \: 1 :- }}$

Find limiting reagent

• For 4 moles of $\bf NH_{3}$ , 5 moles of $\bf O_{2}$ is needed

• for 68 g $\bf NH_{3}$ , 160 g $\bf O_{2}$ is needed

• For 51 g $\bf NH_{3}$ , $\frac{160 × 51 }{68}$ g $\bf O_{2}$ is needed

• For 51 g $\bf NH_{3}$ , $\bf 120$ g $\bf O_{2}$ is needed

• but here , only 48 g is provided so ,

• Limiting reagent = $\bf O_{2}$

$\bf\large\underline{\underline{ Step \: 2 :- }}$

Find amount of $H_{2} O$

• 5 mol $\bf O_{2}$ $\red{\implies}$ 6 mol $H_{2} O$

• 160 g $\bf O_{2}$ $\red{\implies}$ 108 g $H_{2} O$

• 48 g $\bf O_{2}$ $\red{\implies}$ $\frac{108 × 48 }{160 }$

• $\red{\implies}$ 32.4 g $H_{2} O$

$\rule{15cm}{0.02cm}$

ie, 32.4 g $H_{2} O$ is formed