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34kurt
3 years ago
13

A vessel contains a mixture of gases. The mass of each gas used to make the mixture is known. Which of the following information

is needed to determine the mole fraction of each gas in the mixture? The molar mass of each gas A The density of the gases in the vessel B The total pressure of the gases in the vessel C The number of atoms per molecule for each gas
Chemistry
1 answer:
faltersainse [42]3 years ago
8 0

Answer: The molar mass of each gas

Explanation:

Mole fraction is the ratio of moles of that component to the total moles of solution. Moles of solute is the ratio of given mass to the molar mass.

\text{Mole fraction of solute}=\frac{\text{Moles of solute}}{\text{Total Moles}}

Suppose if there are three gases A, B and C.

a) \text{Mole fraction of A}=\frac{\text{Moles of A}}{\text{ Moles of (A+B+C)}}

b) \text{Mole fraction of B}=\frac{\text{Moles of B}}{\text{ Moles of (A+B+C)}}

c) \text{Mole fraction of C}=\frac{\text{Moles of C}}{\text{ Moles of (A+B+C)}}

moles of solute =\frac{\text {given mass}}{\text {Molar mass}}

Thus if mass of each gas is known , we must know the molar mass of each gas to know the moles of each gas.

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2N2H4+ N2O4———3N2+4H2O SalmaKhan99 avatar How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2 Kindly
Alona [7]

Answer:

131.26 g

Explanation:

From the balanced equation,

2 moles of N₂H₄reacts with 1 mole of N₂O₄ to give 3 moles of N₂

Now number of moles of N₂H₄ present in 100 g N₂H₄ is n = 100 g/molar mass N₂H₄.

Molar mass N₂H₄ = 2 × 14.01 g/mol + 1 × 4 g/mol = 28.02 g/mol + 4 g/mol = 32.02 g/mol

n₁ = 100/32.02 = 3.123 mol

Also

Now number of moles of N₂O₄ present in 200 g N₂O₄ is n = 200 g/molar mass N₂O₄.

Molar mass N₂O₄ = 2 × 14.01 g/mol + 16 × 4 g/mol = 28.02 g/mol + 64 g/mol = 92.02 g/mol

n₂ = 200/92.02 = 2.173 mol

Since the mole ratio of N₂H₄  to N₂O₄ is 2 : 1, We require 2 × 2.173 mol N₂H₄  to react with 2.173 mole N₂O₄  

Number of moles of N₂H₄ required is 4.346. But the number of moles of N₂H₄  present is 3.123 so N₂H₄  is the limiting reagent.

So, from the equation, 2 moles of N₂H₄ produces 3 moles of N₂

Therefore number of mole N₂ = 3/2 moles of N₂H₄ = 3/2 × 3.123 mol = 4.6845 mol

From n = m/M where n = number of moles of nitrogen gas = 4.6845 mol and M = molar mass of nitrogen gas = 28.02 g/mol and m = mass of nitogen gas.

m = nM = 4.6845 mol × 28.02 g/mol = 131.26 g

So the mas of nitrogen gas produced is 131.26 g

4 0
3 years ago
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