Question #f69cb

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Question #f69cb

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Answer 1 · 2017-11-07T05:27:04

Answer: 42.58 g of $N H_{3}$ $NH_3$

Explanation:

To solve this question, we need to know the number of moles equivalent to the 35 g of $N_{2}$ $N_2$ using moles concept.

1. Convert 35 g of $N_{2}$ $N_2$ into the number of mols:

Molar mass of $N_{2}$ $N_2$ = 28.0134 g/mol

Therefore,

number of moles = $\frac{35 g}{28.0134 g/mol} = 1.246 m o l s \approx 1.25 m o l s$ $frac(35 g)(28.0134 "g/mol") = 1.246 mols approx 1.25 mols$

We have given the balanced chemical equation as:

$N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 rightarrow 2NH_3$

From above balanced chemical reaction, we get that:

1 mol of $N_{2}$ $N_2$ produces 2 mols of $N H_{3}$ $NH_3$ .

Therefore,

1.25 mols of $N_{2}$ $N_2$ produces 2.5 mols of $N H_{3}$ $NH_3$ .

2. Convert 2.5 mols of $N H_{3}$ $NH_3$ into g:

Molar mass of $N H_{3}$ $NH_3$ = 17.031 g/mol which means

1 mol of $N H_{3}$ $NH_3$ = 17.031 g of $N H_{3}$ $NH_3$
2.5 mols of $N H_{3}$ $NH_3$ = 42.5775 g of $N H_{3}$ $NH_3$ $\approx$ $approx$ 42.58 g of $N H_{3}$ $NH_3$

Answer 2 · 2017-11-07T05:52:13

$4.25 g r a m s$ $4.25grams$

Explanation:

For this first we have to calculate moles of $N_{2}$ $N_2$ from which we can calculate moles of $N H_{3}$ $NH_3$ . And from moles of $N H_{3}$ $NH_3$ , we can easily calculate its mass.
So let's start:

Moles of $N_{2} =$ $N_2 =$ $\frac{m a s s o f N_{2} (g r a m s)}{m o l a r m a s s o f N_{2}}$ $(mass of N_2(grams))/ (molar mass of N_2)$

$= \frac{35 g r a m s}{28 g r a m p e r m o l}$ $= (35grams)/(28grampermol)$ $= 1.25 m o l s$ $= 1.25mols$

Now we have to use balance chemical equation:

$N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 to 2NH_3$

As we can see;

$1 m o l$ $1mol$ $N_{2}$ $N_2$ form number of moles of $N H_{3} = 2$ $NH_3 = 2$

$1.25 m o l s$ $1.25 mols$ $N_{2}$ $N_2$ will form number of moles of $N H_{3}$ $NH_3$

$=$ $=$ $\frac{2 m o l N H_{3}}{1 m o l N_{2}}$ $(2molNH_3)/(1molN_2)$ $\times$ $xx$ $1.25 m o l N_{2}$ $1.25molN_2$

$= 2.5 m o l s N H_{3}$ $= 2.5 molsNH_3$

Mass of $N H_{3} =$ $NH_3 =$ number of moles $\times$ $xx$ molar mass

$= 2.5 \times 17 = 42.5 g r a m s$ $= 2.5 xx 17 = 42.5grams$

$N o t e :$ $Note:$
The coefficients of reactants and products in balance chemical equation represent number of moles. And if there is no coefficient with a molecule then it means that number of mole = 1.

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