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Lelu [443]
2 years ago
10

A sample of N2(g) was collected over water at 25°C and 730 torr in a container with a volume of 340 mL. The vapor pressure of wa

ter at 25°C is 23.76 torr. What mass of N2 was collected?
A. 0.58 g
B. 0.72 g
C. 0.91 g
D. 0.36 g
E. 0.080 g
Chemistry
1 answer:
AnnZ [28]2 years ago
6 0

Answer:

D. 0.36 g

Explanation:

When a gas is collected over water, the total pressure is the sum of the pressure of the gas and the pressure of the water vapor.

Ptotal = Pwater + PN₂

PN₂ = Ptotal - Pwater = 730 torr - 23.76 torr = 706 torr

We can find the mass of N₂ using the ideal gas equation.

P.V=n.R.T=\frac{m}{M} .R.T\\m=\frac{P.V.M}{R.T} =\frac{730torr.0.340L.28g/mol}{(0.08206atm.L/mol.K).298K} .\frac{1atm}{760torr} =0.36g

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When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain
yKpoI14uk [10]

<u>Answer:</u>

<u>For a:</u> The number of molecules of nitrogen dioxide is 4.52\times 10^{23}

<u>For b:</u> The mass of nitric acid formed is 54.81 grams

<u>For c:</u> The mass of nitric acid formed is 206 grams

<u>Explanation:</u>

The given chemical reaction follows:

3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)

  • <u>For a:</u>

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with \frac{3}{1}\times 0.250=0.75mol of nitrogen dioxide

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.75 moles of nitrogen dioxide will contain 0.75\times 6.022\times 10^{23}=4.52\times 10^{23} number of molecules

Hence, the number of molecules of nitrogen dioxide is 4.52\times 10^{23}

  • <u>For b:</u>

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

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

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = \frac{2}{3}\times 1.304=0.870 moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g

Hence, the mass of nitric acid formed is 54.81 grams

  • <u>For c:</u>
  • <u>For nitrogen dioxide:</u>

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol

  • <u>For water:</u>

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = \frac{1}{3}\times 4.90=1.63mol of water

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

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

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce \frac{2}{3}\times 4.90=3.27mol of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g

Hence, the mass of nitric acid formed is 206 grams

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2 years ago
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