Question

The following reaction was performed in a 1.40 L non-leaking container at a constant temperature of 149.2 ∘C.

C₃H₈(g)+3H₂O(g)→3CO(g)+7H₂(g)

Upon completion of the reaction 0.410 g H2 was formed.

Part A

What was the pressure of H2 after the reaction was completed?

Part B

What was the minimum starting pressure of C3H8?

Part C

What was the minimum starting pressure of H2O?

Part D

If stoichiometric amounts of C3H8 and H2O were used and the reaction went to completion, what was the total pressure of the system before the reaction took place?

Part E

If stoichiometric amounts of C3H8 and H2O were used, what was the total pressure of the system after the reaction went to completion?

Answer 1

V=1.40L T = 149.2°C = 149.2°C + 273.15 = 422.35 K (3 Hg (9) + 3 H2O (9) 3 CO (9) + 7 H2 (9) Upon completion 0.410g of H2 was formed ПРАрин өн, 1 касиет шол completed moles: Cras constant PV = nRT = P = n RT Temp. Vy volume Pa 0.410 x 6-0821 L atm. mol ) x 422:35 ] P = 0.410 x H 544 24. 76 - NINSKI m 5.077 atm. 1.40L 2 Minimum starting pressure of C3H8 1 Acc. to Balanced chemical equation. 7 moles of H₂ were produced when Imole of (3 H8 were was Se 10. 205 " " " " 1 0.205 mole of 7 (3 Hs were use moles of H2 = 0.410 0.205 = 0.029 mole of C3H6.

DATE: / PAGE NO 20 Minimum starting pressure of C3H8 P = nRT v = 0.029 X 0.0821 1.40 422.35 = 0.029 x 24.76 = 0.725 atm (c) Minimum starling presswel of H2O Acc. to balanced chemical equation moles of H₂ were produced when 3 mole of H₂O щщщed 0.205 1 " " 11 11 3 x 0.205 mole of H₂0 wey used 0.087 mole of H₂O = P = NRT = 0.087 X 24.76 - 6. .154 atm Total pressue of system before reaction take place = P GHS + PH2O = 0.725 + 2.154 2.879 abm

Total Pressure of the syslem after reaction went to completion P= 2.87X moles of product moles of React = 2.87 X (3+1) = 2.87&tos Since Stoichiomet amount of (BHL . HD a 'ed, we have taken their stiochiometr cofficient in calculating moles 42 Teaben 7.17 atm