Question

If a reaction vessel contains 51 grams of NH3 what is maximum mass of HCN that could be made assuming the reaction goes to completion? Find also the volume of O2 And CH4 gas's required the reaction at S.T.P

Answer 1

Answer:

The reaction at STP requires around 49.06 L of O2 gas and 16.91 L of CH₄ gas.

Explanation:

The reaction between NH₃ and O₂ to produce HCN and H₂O has the following chemical equation:

$4NH_3 + 3O_2 → 2HCN + 6H_2O$

The balanced chemical equation states that two moles of HCN are created when four moles of NH₃ and three moles of O₂ are combined. NH₃ has a molar mass of 17.03 g/mol, while HCN has a molar mass of 27.03 g/mol. Consequently, 3 moles of NH₃ are equal to 51 g of NH₃.

The mole ratio from the balanced chemical equation can be used to calculate the maximum mass of HCN that could be produced assuming the reaction proceeds to completion:

$( \frac{2 \: moles \: HCN }{ 4 \: moles \: NH_3}) \times (27.03 g/mol \: HCN) \times (3 \: moles NH_3)$

= 40.54 g HCN

As a result, the largest mass of HCN that may be produced is roughly 40.54 g.

We may use the ideal gas law to determine the volume of O₂ and CH₄ gases needed for the reaction at STP:

PV = nRT

where R is the gas constant, n is the number of moles, P is the pressure, V is the volume, and T is the temperature. The temperature and pressure are both 1 atm at STP.

The balanced chemical equation can be used to determine how many moles of oxygen are necessary:

$( \frac{3 \: moles \: O_2 }{ 4 \: moles \: NH_3}) \times (3 \: moles \: NH_3) = 2.25 \: moles \: O_2$

The following formula can be used to determine the necessary amount of O2:

$V = \frac{(nRT) }{ P}$

$= ( \frac{2.25 mol \times 0.08206 L atm K^{-1} mol^{-1} \times 273 K) }{ 1 atm}$

≈ 49.06 L

Similarly, using the balanced chemical equation, it is possible to determine how many moles of CH4 are needed:

$( \frac{1 \: mole \: CH_4 }{ 4 \: moles \: NH_3}) \times (3 \: moles \: NH_3) = 0.75 \: moles \: CH_4$

Calculate the necessary volume of CH₄ as follows:

$V = \frac{(nRT) }{ P}$

$= \frac{(0.75 mol \times 0.08206 L atm K^{-1} mol^{-1} \times 273 K) }{1 atm}$

≈ 16.91 L

As a result, the reaction at STP requires around 49.06 L of O2 gas and 16.91 L of CH₄ gas.

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