Asked by Gweedo
The problem:
Nitrogen can react with steam to form ammonia and nitrogen monoxide gas. A 20.0L sample of nitrogen at 173C and 772mmHg is made to react with an excess of steam. The products are collected at room temperature (25C) into an evacuated flask with a volume of 15.0
A. Write a balanced equation for the reaction.
B. What is the total pressure of the products in the collecting flask after the reaction is complete?
C. What is the partial pressure of each of the products in the flask?
My homework:
A. N(g) + H2O(g)--> NO(g)+ NH3(g)
I also determined that:
T=173C=446.15K & 25C=298.15
P=772mmHg=1.016atm
I believe I have A. correct but B and C are giving me a problem. I am sure I could get it after some sleep but I have been doing chemistry for 10 hours straight and could really use a little more help on this last problem.
Nitrogen can react with steam to form ammonia and nitrogen monoxide gas. A 20.0L sample of nitrogen at 173C and 772mmHg is made to react with an excess of steam. The products are collected at room temperature (25C) into an evacuated flask with a volume of 15.0
A. Write a balanced equation for the reaction.
B. What is the total pressure of the products in the collecting flask after the reaction is complete?
C. What is the partial pressure of each of the products in the flask?
My homework:
A. N(g) + H2O(g)--> NO(g)+ NH3(g)
I also determined that:
T=173C=446.15K & 25C=298.15
P=772mmHg=1.016atm
I believe I have A. correct but B and C are giving me a problem. I am sure I could get it after some sleep but I have been doing chemistry for 10 hours straight and could really use a little more help on this last problem.
Answers
Answered by
GK
The chemical equation is:
5N2(g) + 6H2O(g)--> 6NO(g) + 4NH3(g)
The coefficients can be used either as a mole ratio or a volume ratio.
The volumes of the products are:
20.0L of N2 x (6.00 L NO / 5 L N2) = 24.0 L NO
20.0L of N2 x (4.00 L NH3 / 5 L N2) = 16.0 L NH3
TOTAL volume of products = 24.0L + 16.0L = 40.0L
The mole fractions of each product gas are:
X(NO) = 24/40 = 0.600
X(NH3) = 16/40 = 0.400
• To find the final total pressure in the flask use:
PV = nRT
• To find the final pressures of NO and NH3, multiply each mole fraction by the previous answer.
I am letting you do the last two part of the solution on your own.
5N2(g) + 6H2O(g)--> 6NO(g) + 4NH3(g)
The coefficients can be used either as a mole ratio or a volume ratio.
The volumes of the products are:
20.0L of N2 x (6.00 L NO / 5 L N2) = 24.0 L NO
20.0L of N2 x (4.00 L NH3 / 5 L N2) = 16.0 L NH3
TOTAL volume of products = 24.0L + 16.0L = 40.0L
The mole fractions of each product gas are:
X(NO) = 24/40 = 0.600
X(NH3) = 16/40 = 0.400
• To find the final total pressure in the flask use:
PV = nRT
• To find the final pressures of NO and NH3, multiply each mole fraction by the previous answer.
I am letting you do the last two part of the solution on your own.
Answered by
Gweedo
Please disregard I finally figured it out.
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