27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams

Question

3 years ago

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27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams

of the excess reagent remain after the reaction? Hint: Write a balanced chemical equation for the reaction first. Enter your answer to 1 decimal place.

Chemistry

1 answer:

GarryVolchara [31] · Answer 1 · 2022-01-15T15:43:59+03:00

Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aluminium nitrite:

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol$

For ammonium chloride:

Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol$

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

$Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = $\frac{3}{1}\times 0.668=2.004mol$ of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g$

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams