Answer to Question #184993 in Inorganic Chemistry for julia

Question #184993

A reaction to convert ammonia into nitric acid involves following chemical

equation:

4 NH3 (g) + 5 O2 (g)à 4 NO(g) + 6 H2O(g)


Assume that 1.50 g of NH3 reacts with 2.75 g O2, then:

(a) Determine the limiting reactant.


(b) Calculate how many grams of NO and H2O will form in the reaction above.


1
Expert's answer
2021-07-06T00:59:58-0400

Moles of Ammonia = "\\frac{1.50}{17}=0.088 mol"


Moles of Oxygen"=\\frac{2.75}{32}=0.086 mol"



"Difference=0.088\u22120.086=2.0\u00d710^{\u22123}mol"


Mass of excess reactant that remain = "[2.0\u00d710^{-3}]\u00d717=0.034grams"



Moles of NO "=\\frac{1.80}{30}=0.06 moles"


Moles of H2O="=\\frac{1.50}{33}=0.045 moles"


Limiting reactants is NH3

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Comments

ANIQ
02.07.21, 18:46

Where is the answer for (b) question?

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