Question

From 51 gm of NH3 and 90 gm of H2O all Hydrogen is removed and converted to Hydrogen gas. Find mass and moles of hydrogen gas.

Answer 1

Concept Introduction: Stoichiometric Chemistry is the basic concept of Chemistry.

Given:

We have been Given: mass of NH3 given

$51g$

mass of H2O given

$90g$

To Find:

We have to Find: Find the moles of Hydrogen removed.

Solution:

According to the problem, Removing hydrogen from Ammonia is

$\frac{51}{14 +(3 \times 1)} = \frac{51}{17} = 3 \: moles$

Removing hydrogen from water is,

$\frac{90}{(2 \times 1) + 16} = \frac{90}{18} = 5 \: moles$

therefore total removed hydrogen is,

$3 + 5 = 8 \: moles$

Final Answer: Total Removed hydrogen is

$8 \: moles$

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Answer 2

Answer:

3 moles of NH₃ gives $\frac{9}{2}$ moles of H₂, i.e., 4.5 g of H₂.

5 moles of H₂O gives $\frac{5}{2}$ moles of H₂, i.e., 2.5 g of H₂.

Explanation:

We are given 51 g of NH₃ and 90 g of H₂O. Now, if all the hydrogen is removed and converted into a gaseous form, we have to find the mass of hydrogen and the number of moles formed.

To do, this we need to first calculate the number of moles of NH₃ and H₂O that we have.

We know that the molar mass of NH₃ is 17 g and that of H₂O is 18 g.

Thus, when given 51 g of NH₃, we can calculate it to be 3 moles of NH₃.

Also, when given 90 g of H₂O, we can calculate it to be 5 moles of H₂O.

We also know the decomposition reactions of both these compounds to be as follows -

$2NH_3 \rightarrow N_2 + 3H_2\\\implies 3NH_3 \rightarrow \frac{3}{2} N_2 + \frac{9}{2} H_2$

Thus, 3 moles of NH₃ gives $\frac{9}{2}$ moles of H₂, i.e., 4.5 g of H₂.

$2H_2O \rightarrow 2H_2 + O_2\\\frac{5}{2} H_2O \rightarrow \frac{5}{2}H_2 + \frac{5}{4} O_2$

Thus, 5 moles of H₂O gives $\frac{5}{2}$ moles of H₂, i.e., 2.5 g of H₂.

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