Question

3) Calculate the grams of CO2 that would be produced from the complete burning of 18.2 L of liquid ethanol, C HOH, measured at 0.823 atm and 27.0°C. CHOH (I) + 3 0: (g) → 2 CO2(g) + 3 H2O(g) 4) Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4H2O(g) + 3Fe(s) - Fe,O4(s) + 4H2(g) What volume of Ha(g) at a pressure of 735 torr and a temperature of 19.0 °C can be prepared from the reaction of 35.9 g of Fe 3) In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 743 torr and a temperature of 29.0 °C. The mass of the gas was 0.472 g. a) How many moles of gas were in the sample? Water vapor pressure at this temperature is 30.0 torr (subtract this first from 743 torr first)

Answer 1

Ans 3 Given: Volume, V = 18.22, Pressure, P = 0·823 atm. Temperature, I = 27.0°C = 300 k. Universal gas constant, R = 0.0821 atm L/k/mol. Using ideal gas equation, PV = nRT n = PV = 0.823 atm x 18.2L RT 0.082) atm L/k/mol X 3ook => = 0.608 mol Now, according to balanced chemical reaction, produces 2 mol of coa 1 mol of SHOH i Moles of coa produced H5OH = 2x moles of 22x0.608 mol = 1-216 mol Molar mass And, Mass of co = Moles of cox = 1.216 X 44 g = 53.5049.

4. According to the balanced chemical equation, 3 mol of Fe produces 4 mol of H₂.

Molar mass of Fe = 56 g/mol

Molar mass of H₂ = 2 g/mol

35.9 g Fe will be equal to 35.9 g ÷ 56 g/mol = 0.64 mol.

So, using unitary method, 3 mol Fe produces 4 mol of H₂

=> 1 mol of Fe will produce 4/3 mol of H₂

And, 0.64 mol of Fe will produce (4/3) × 0.64 mol of H₂.

= 0.85 mol of H₂.

(Alternate method for calculating number of moles of Hydrogen gas:

So, 3 mol × 56 g/mol Fe produces 4 mol × 2 g/mol of H₂.

=> 168 g Fe produces 8 g of H₂ gas.

Now, using unitary method

1 g of Fe produces 8/168 g of H₂.

So, 35.9 g of Fe produces (8/168) × 35.9 g of H₂.

= 1.71 g of H₂ will be produced.

Number of moles of H₂ = 1.71 g ÷ 2 g/ mol = 0.85 mol).

Now, use ideal gas equation, PV = nRT.

P = 735 torr = 0.967 atm

T = 19 ℃ = 292 K

n = 0.85 mol (calculated above)

R = 0.0821 L.atm/K.mol

nRT V =

=>

v_0.85 x 0.0821 x 292 0.967

=> V = 21.07 L.

5.

Total pressure = pressure of the gas + pressure of water vapour

=> 743 torr = pressure of the gas + 30 torr

=> Pressure of the gas = 743 - 30 torr = 713 torr = 0.94atm.

Volume of the gas, V = 265 mL = 0.265 L

Temperature, T = 29 ℃ = 302 K

Using ideal gas equation, PV = nRT

PI n = RT

=>

n = 0.94 x 0.265 0.0821 x 302

=> n = 0.01 mol.