Help, please show and explain full answers in order to use as a future refrence

Question

Help, please show and explain full answers in order to use as a future refrence for tests!

1. To find the pressure of oxygen the vapor pressure of water has to be considered, why?

2. Why are the water levels inside the flask and the beaker equalized before starting the experiment?

3. A certain mass of magnesium is reacted with 6M HCl. The hydrogen gas liberated is collected by displacing 200 mL of water. The pressure of the gas and the atmospheric pressure are equalized. The atmospheric pressure is 762 mmHg. The temperature of the experiment is 27oC. How many moles of hydrogen have been liberated, and what is the mass of the magnesium that reacted?

4. A student forgot to take the water vapor pressure into consideration. How would that effect the number of moles of oxygen and what effect would it have on the value of the gas constant R? Explain.

5. You are given 1.56 g of a mixture of KClO3 and KCl. When the mixture is heated the oxygen gas liberated displaces 327 mL of water at a temperature of 19oC. The total pressure of the gas in the collection flask is 735 mmHg. What is the percentage of KClO3 in the mixture?

Explanation / Answer

1. The total pressure of the system is the sum total of pressure exterted by all the gases in the vessel at that time. Since water vapor is in equilibrium with the liquid water, the pressure of oxygen is calculated by substracting the vapor pressure of water at that temperature from the total pressure measured at that time.

2. The level of water inside the flask and the beaker are equalized as the pressure in the system must be equalized to the atmospheric pressure at that temperature.

3. pressure of H2 collected = 762 - 26.7 = 735.3 mmHg

Volume = 0.2 L

Temperature = 273 + 27 = 300 K

So,

moles of H2 collected = (735.3/760)(0.2)/(0.08205)(300)

                                    = 0.008 mol

Mg + 2HCl --> MgCl2 + H2

mass of Mg reacted = 0.008 mol x 24.305 g/mol = 0.194 g

4. If pressure of water vapor is not substracted, the moles of H2 collected would be higher than actual value and thus, value of R would be lower than the actual value.

5. Pressure of O2 = 735 - 16.5 = 718.5 mmHg

Volume = 0.327 L

Temperature = 273 + 19 = 292 K

So,

moles O2 = (718.5/760)(0.327)/(0.08205)(292) = 0.013 mol

2KClO3 --> 2KCl + 3O2

moles KClO3 = 0.013 x 2/3 = 0.0087 mol

mass KClO3 = 0.0087 mol x 122.55 g/mol = 1.06 g

Percentage KClO3 in the sample = 1.06 x 100/1.56 = 68.10%

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