Asked by Juan
we burn 12.50L of ammonia in 20.00L of oxygen at 500 degrees celsius. What volume of nitric oxide, NO, gas can form? what volume of steam, H2O (g) is formed? Assume that all gases are at the same temperature and pressure, and that the limiting reactant is used up
4NH3(g) + 5O2(g) =====> 4NO(g) + 6H2O(g)
not really sure what to do
4NH3(g) + 5O2(g) =====> 4NO(g) + 6H2O(g)
not really sure what to do
Answers
Answered by
Anon
use your ideal gas formula, but first find the limiting reagent to find the moles of NO. Then to find volume
V=(nRT)/P assume pressure to be one and temperature to be 273K R is the constant .08206. Hoped it helped. The total volume is 12.50+20 (this is true because of the law of conservation of mass) What ever volume you get for NO subtract it from the total to find the H2O volume.
V=(nRT)/P assume pressure to be one and temperature to be 273K R is the constant .08206. Hoped it helped. The total volume is 12.50+20 (this is true because of the law of conservation of mass) What ever volume you get for NO subtract it from the total to find the H2O volume.
Answered by
Anon
n is the moles of No
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