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Question Number 17179 by tawa tawa last updated on 01/Jul/17

	Answered by Tinkutara last updated on 02/Jul/17
	$Q.1. Moles of H_2 = ((10)/2) = 5 N_2 = ((50)/(28)) ≈ 1.786 CO_2 = ((40)/(44)) ≈ 0.9 Average Molecular weight = ((100)/(5 + ((25)/(14)) + ((10)/(11)))) ≈ 12.9958 Q.2. Moles of N_2 = ((4000)/(28)) = ((1000)/7) Moles of CO_2 = ((6000)/(44)) = ((1500)/(11)) Total number of moles = ((21500)/(77)) Mole fraction of N_2 = ((1000)/7)×((77)/(21500)) ≈ 0.512 Mole fraction of CO_2 = ((1500)/(11))×((77)/(21500)) ≈ 0.488 Average molecular weight = 10000×((77)/(21500)) ≈ 35.814 Partial pressure of N_2 = 0.512 × 4 bar = 2.048 bar Partial pressure of CO_2 = 0.488 × 4 bar = 1.952 bar$
	$Q . 1 . Moles of H_{2} = \frac{10}{2} = 5$ $N_{2} = \frac{50}{28} \approx 1.786$ ${CO}_{2} = \frac{40}{44} \approx 0.9$ $Average Molecular weight$ $= \frac{100}{5 + \frac{25}{14} + \frac{10}{11}} \approx 12.9958$ $Q . 2 . Moles of N_{2} = \frac{4000}{28} = \frac{1000}{7}$ $Moles of {CO}_{2} = \frac{6000}{44} = \frac{1500}{11}$ $Total number of moles = \frac{21500}{77}$ $Mole fraction of N_{2} = \frac{1000}{7} \times \frac{77}{21500}$ $\approx 0.512$ $Mole fraction of {CO}_{2} = \frac{1500}{11} \times \frac{77}{21500}$ $\approx 0.488$ $Average molecular weight$ $= 10000 \times \frac{77}{21500} \approx 35.814$ $Partial pressure of N_{2} = 0.512 \times 4 bar$ $= 2.048 bar$ $Partial pressure of {CO}_{2} = 0.488 \times 4 bar$ $= 1.952 bar$
	Commented by tawa tawa last updated on 02/Jul/17

	$Wow, i really appreciate sir . God bless you sir .$
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