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GenaCL600 [577]
3 years ago
6

A mixture of nitrogen gas and oxygen gas has a total pressure of l.13 atm. If the partial pressure of oxygen is 0.43 atm, what i

s the partial pressure of nitrogen?
a) 1.56 atm
b) 1.13 atm
c) 0.70 atm
d) 0.38 atm​

Chemistry
2 answers:
Vladimir [108]3 years ago
8 0
C
0.70

I hope this is help, I’m so so sorry if I’m incorrect
Stels [109]3 years ago
6 0

Answer:

0.70 atm - Option C

Explanation:

Let's apply the mole fraction concept to solve this:

Partial pressure of a gas / Total pressure = Mole fraction of a gas.

In a mixture, sum of mole fraction from each gas = 1

Partial pressure of O₂ / Total pressure = Mole fraction of O₂

0.43 atm / 1.13 atm = 0.380

1 - 0.380 = 0.62 → Mole fraction N₂

Mole fraction N₂ = Partial pressure N₂ / Total pressure

0.62 = Partial Pressure N₂ / 1.13 atm

0.62 . 1.13 = 0.70 atm → Partial Pressure N₂

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<u>Answer:</u> The mass of cryolite produced is 51.48 kg

<u>Explanation:</u>

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

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

  • <u>For aluminium oxide:</u>

Given mass of aluminium oxide = 12.5 kg = 12500 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol

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Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol

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Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol

For the given chemical reaction:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with (6\times 122.6)=735.6mol of sodium hydroxide and (12\times 122.6)=1471.2mol of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

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

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = \frac{2}{1}\times 122.6=245.2mol of cryolite

Now, calculating the mass of cryolite by using equation 1:

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, 51477.3 g\times (\frac{1kg}{1000g})=51.48kg

Hence, the mass of cryolite produced is 51.48 kg

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