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For the following reaction, 20.4 grams of nitrogen monoxide are allowed to react with 5.10 grams of hydrogen gas . nitrogen monoxide(g) hydrogen(g) nitrogen(g) water(l) What is the maximum amount of nitrogen gas that can be formed

User Adey
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1 Answer

7 votes

Answer:

9.52g of N2

Step-by-step explanation:

The equation of the reaction is given as;

2NO(g) + 2H2(g) → N2(g) + 2H2O(l)

From the stochiometry of the reaction;

2 mol : 2 mol : 1 mol : 2 mol

60g : 4g : 28g :

( Mass = Number of moles * Molar mass)

Converting the masses given to moles;

Number of moles of NO = mass / molar mass = 20.4 / 30 = 0.68

Number of moles of H2 = mass / molar mass = 5.10 / 2 = 2.55

In this reaction, the limiting reactant is NO

60g of NO produces 28g of N2,

20.4g of NO would produce xg of N2

60 = 28

20.4 = X

X = (20.4 * 28) / 60

X = 9.52g

User Aquilesb
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