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miskamm [114]
3 years ago
8

If you have 2.1 liters of hydrogen gas (at STP), what mass of hydrogen gas would you have?

Chemistry
1 answer:
GuDViN [60]3 years ago
6 0

Answer:

0.19 g

Explanation:

Step 1: Given data

Volume of hydrogen at standard temperature and pressure (STP): 2.1 L

Step 2: Calculate the moles corresponding to 2.1 L of hydrogen  at STP

At STP (273.15 K and 1 atm), 1 mole of hydrogen has a volume of 22.4 L if we treat it as an ideal gas.

2.1 L × 1 mol/22.4 L = 0.094 mol

Step 3: Calculate the mass corresponding to 0.094 moles of hydrogen

The molar mass of hydrogen is 2.02 g/mol.

0.094 mol × 2.02 g/mol = 0.19 g

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Calculate the number of moles and the mass of the solute in each of the following solutions:
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<u>Answer:</u>

<u>For a:</u> The number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

<u>For b:</u> The number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

<u>For c:</u> The number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

<u>For d:</u> The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}    .....(1)

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(2)

  • <u>For a:</u>

Molarity of KI = 8.23\times 10^{-5}M

Volume of solution = 325 mL = 0.325 L     (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol

Now, using equation 2, we get:

Moles of KI = 2.7\times 10^{-5}mol

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g

Hence, the number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

  • <u>For b:</u>

Molarity of sulfuric acid = 22\times 10^{-5}M

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol

Now, using equation 2, we get:

Moles of sulfuric acid = 1.65\times 10^{-5}mol

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g

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Molarity of potassium chromate = 0.1135M

Volume of solution = 0.250 L

Putting values in equation 1, we get:

0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol

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Moles of potassium chromate = 2.84\times 10^{-2}mol

Molar mass of potassium chromate = 194.2 g/mol

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2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g

Hence, the number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

  • <u>For d:</u>

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

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Putting values in equation 2, we get:

39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g

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