Provide complete solutions for full credit. Please place your final answers in t

Question

Provide complete solutions for full credit. Please place your final answers in the boxes provided. What volume (in liters) of N_2O_5 gas 25.0 degree C and 0.768 atm is required to produce 200 g of LiNO_3 by its reaction with excess Li_2O as shown below, assuming a reaction yield of 100% N_2O_3(g) + Li_2O(s) rightarrow 2 LiNO_3(s) A 1.0 L flask containing 140.0 g of SO_3 is allowed to come to equilibrium at 850 degree C via the reaction: 2So_3(g) O_2(g) + 2SO_2(g) K = 5.7 times 10^6 (at 850 degree C) How many grams of SO_2 are produced upon reaching equilibrium?

Explanation / Answer

Question 11

Balanced equation:
N2O5(g) + Li2O(s) = 2 LiNO3(s)

200 gm LiNO3 = 200/ 68.94 = 2.9 mol

to get 2.9 mol of LiNO3 we should have 1.45 mol of N2O5 according to the above equation.

Let us calculate the volume of N2O5

PV =nRT

P = 768 mmHg = 1.01 atm

V= ?

n= 1.45 mol

R = gas constant = 0.0821 L atm K-1 mol-1

T = 298 K

V = 1.45 x 0.821 x 298 / 1.01 = 35.12 Liter

Hence 35.12 liter of N2O5 have been used

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