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olga_2 [115]
2 years ago
10

A self-contained underwater breathing apparatus (SCUBA) uses canisters containing potassium superoxide. The superoxide consumes

the CO2 exhaled by a person and replaces it with oxygen. 4 KO2(s) + 2 CO2(g) n 2 K2CO3(s) + 3 O2(g) What mass of KO2, in grams, is required to react with 8.90 L of CO2 at 22.0 °C and 767 mm Hg
Chemistry
1 answer:
Mrac [35]2 years ago
8 0

Answer:

52.0004 grams of mass of potassium superoxide  is required

Explanation:

Let moles carbon dioxide gas be n at 22.0 °C and 767 mm Hg occupying 8.90 L of volume.

Pressure of the gas,P = 767 mm Hg = 0.9971 atm

Temperature of the gas,T = 22.0 °C = 295.15 K

Using an ideal gas equation to calculate the number of moles.

PV=nRT

n=\frac{0.9971 atm\times 8.90 L}{0.0821 atm L/mol K\times 295.15 K}

n = 0.3662 mol

4KO_2(s)+2CO_2(g)\rightarrow 2K_2CO_3(s)+3O_2(g)

According to reaction, 2 moles of carbon-dioxide reacts with 4 moles of potassium superoxide.

Then 0.3662 mol of  of carbon-dioxide will react with:

\frac{4}{2}\times 0.3662 mol=0.7324 mol of potassium superoxide.

Mass of 0.7324 mol potassium superoxide:

0.7324 mol × 71 g/mol = 52.0004 g

52.0004 grams of mass of potassium superoxide is required.

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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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869.6 × 10¹⁴ molecules of EDB

Explanation:

We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.

We need to transform lb in grams.

1 lb = 453.6 grams

1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams

Now we determine the number of molecules of EDB in the sample by devise the following reasoning:

if we have        31.5 × 10⁻⁹ g of EDB in 1 g of sample

then we have   X  g of EDB in 861.8 g of sample

X = (31.5 × 10⁻⁹ × 861.8) / 1 = 27146.7 × 10⁻⁹ g of EDB

Molecular mass of EDB (C₂H₄Br₂) = 188 g/mole

Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:

if       188 g of EDB contains 6.022 × 10²³ molecules

then 27146.7 × 10⁻⁹ g of EDB contains Y molecules

Y = (27146.7 × 10⁻⁹ × 6.022 × 10²³) / 188 = 869.6 × 10¹⁴ molecules of EDB

Learn more:

about Avogadro’s number

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