5. Hydrogen gas (2.02 g/mol) can be produced from the reaction of methane (16.05

Question

5. Hydrogen gas (2.02 g/mol) can be produced from the reaction of methane (16.05 g/mol) and water vapor (18.02 g/mol) according to the following reaction equation: CHdg) + H2O(g) CO(g) + 3 Hag) If98 g CHs react with excess H20 at 733 K and 0.762 atm, what volume of H2 gas will form? Question 4 5. At a given temperature, nitrogen gas effuses 0.378 times faster than the gas X, what is the molar mass of X? Question5 5. Tungsten hexafluoride is one of the heaviest gases known, with a molar mass of 297.9 g/mol. At a given temperature, how long will it take WFs to cover the distance H2 can travel in 33 seconds? Question 6 5. You wish to measure the changes in pressures of gases during the course of a reaction using a closed-end mercury manometer. Suppose 720 mmHg N204 is allowed to react to form NO2 according to the reaction, N204g)- 2 NO2(g). The total pressure you measure, as the reaction proceeds, is the sum of the pressures of N204 and NO2 present. When the total pressure is 1150 mmHg, what is the pressure of N204? (mmHg) Question 7 5. The diffusion rate of H2 in a lecture hall was found to be 22.0 m/sec at 25°C. How long will it take poisonous HCN gas to diffuse 9 m in the same room? (s) Question 8 5. During the decomposition of CaCO,(s) into Ca0(s) and CO.(g), 115.0 ml of gas is collected by the displacement of water in a flask at 21°C with an atmospheric pressure of 735 mmHg. The vapor pressure of water at this temperature is 18.7 mmHg. What was the mass of oxygen collected? (mg)

Explanation / Answer

CH4 (g) + H2O (g) .......................... CO (g) + 3H2 (g)

From the above equation it is clear that one mole of methane will produce three moles of hydrogen gas.

if 98 g of methane will react then,

First convert 98 g of methane in to its number of moles

= 98/16.05, = 6.106 mole

Therefore, by reacting 6.106 moles of methane in the presence of excess of water, 18.318 mole of H2 gas will be produced.

One mole of a gas at STP occupies 22.4 litres. So at STP the volume occupied by 18.318 moles of H2 gas will be

= 22.4 * 18.318

= 410.323 litres.

Convert this volume at STP to the volume at the given conditions of temperature and pressure by applying the formula

(P1V1)/T1 = (P2V2)/T2

(1 x 410.323)/273.15 = (0.762 x V2)/733

(410.323 x 733)/(273.15 x 0.762) = V2

1445 litres = V2 (Volume of H2 gas produced by 98 g of methane at 733 k and 0.762 atm pressure

Please post rest of the questions separately.

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