4NH3(g) + 7O2(g) --> 4NO2(g) + 6H2O(g) 1) What type of reaction if this? 2) if t

Question

4NH3(g) + 7O2(g) --> 4NO2(g) + 6H2O(g)
1) What type of reaction if this?
2) if the reaction runs 100% completion and 28.8g of ammonia are completely combusted, how many grams of oxygen are consumed?
3) if 4.50 grams of ammonia react with 8.80 grams of oxygen what is the maximum amount of nitrogen dioxide that can be made? 4NH3(g) + 7O2(g) --> 4NO2(g) + 6H2O(g)
1) What type of reaction if this?
2) if the reaction runs 100% completion and 28.8g of ammonia are completely combusted, how many grams of oxygen are consumed?
3) if 4.50 grams of ammonia react with 8.80 grams of oxygen what is the maximum amount of nitrogen dioxide that can be made?
1) What type of reaction if this?
2) if the reaction runs 100% completion and 28.8g of ammonia are completely combusted, how many grams of oxygen are consumed?
3) if 4.50 grams of ammonia react with 8.80 grams of oxygen what is the maximum amount of nitrogen dioxide that can be made?

Explanation / Answer

4NH3 (g) + 7O2 (g) --> 4NO2 (g) + 6H2O (g)

1) Combustion reaction

2) Molar mass of NH3 = 17.031 g/mol

28.8g of Ammonia

Moles of ammonia =28.8g/17.031 g/mol = 1.691 moles

4 moles of NH3 reacts with 7 moles of O2

1.691 moles of NH3 react with 1.691 moles x (7/4) = 2.96 moles of O2

Now convert moles of O2 into g of O2

2.96 moles x 32 g/mol = 94.72 g

94.72 grams of oxygen are consumed.

3) if 4.50 grams of ammonia react with 8.80 grams of oxygen what is the maximum amount of nitrogen dioxide that can be made?

Calculate moles of 4.50 grams of ammonia and 8.80 grams of oxygen

Moles of ammonia = 4.50g/17.031 g/mol = 0.264 moles

Moles of O2 = 8.80g/32g/mol = 0.275 moles

So O2 is a limiting reagent

7moles of O2 produces 4 moles of NO2

So, 0.275 moles of O2 produces0.275 x (4 /7) = 0.157 moles of NO2

Now convert moles of NO2 into g of NO2

0.157 moles of NO2 x 46 g/mol = 7.23 g

7.23 g of nitrogen dioxide that can be made

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