One of the steps in the commercial process for converting ammonia to nitric acid

Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO:

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

In a certain experiment, 1.40g of NH3 reacts with 2.57g of O2.

1))) How many grams of NO and of H2O form?

Enter your answers numerically separated by a comma.

2))) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements: 2NaN3(s)?2Na(s)+3N2(g)

3))) How many moles of N2 are produced by the decomposition of 1.60mol of NaN3?

4))) How many grams of NaN3 are required to form 14.0g of nitrogen gas?

Explanation / Answer

Note: Be careful how you list your data. I can only assume that "2.05" stands for "2.05 g of NH3" and that "3.76 of O2" stands for "3.76 g of O2".
4 NH3(g) + 5 O2(g) ? 4 NO(g) + 6 H2O(g)

First question:
How many grams of NO and of H2O form?
Before we can calculate that answer, we need to know which reactant is the limiting reactant, and which is present in excess. To do this, let's assume that NH3 is the limiting reactant. If so, we need to know whether there is enough O2 to react with all of 2.05 g of NH3. We can use dimensional analysis, and equation coefficients to convert 2.05 g of NH3 to moles of NH3, to moles of O2, to grams of O2. We shall need the following equalities to set up conversion factors:
1 mol NH3 = 17.03052 g NH3
4 mol NH3 = 5 mol O2
1 mol O2 = 31.9988 g O2
[(2.05 g NH3)/1][(1 mol NH3)/(17.03052 g NH3)][(5 mol O2)/(4 mol NH3)][(31.9988 g O2)/(1 mol O2)] = 4.8147026 g O2
Since as much as 4.8147026 g of O2 is needed to react completely with 2.05 g of NH3, and only 3.76 g of O2 is available, then O2 is the limiting reactant, and NH3 is the reactant in excess.
Now we can calculate the number of grams of NO that will be formed from 3.76 g of O2, by using dimensional analysis to convert 3.76 of O2 to moles of O2 to moles of NO, to grams of NO. We shall need the following equalities to set up conversion factors:
1 mol O2 = 31.9988 g O2
4 mol NO = 5 mol O2
1 mol NO = 30.0061 g NO
[(3.76 g O2)/1][(1 mol O2)/(31.9988 g O2)][(4 mol NO)/(5 mol O2)][(30.0061 g NO)/(1 mol NO)] = 2.820684 g NO or 2.82 g NO rounded to three significant figures
Next, we can calculate the number of grams of H2O that will be formed from 3.76 g of O2, by using dimensional analysis to convert 3.76 of O2 to moles of O2 to moles of NO, to grams of H2O. We shall need the following equalities to set up conversion factors:
1 mol O2 = 31.9988 g O2
5 mol O2 = 6 mol H2O
1 mol H2O = 18.01528 g H2O
[(3.76 g O2)/1][(1 mol O2)/(31.9988 g O2)][(6 mol H2O)/(5 mol O2)][(18.01528 g H2O)/(1 mol H2O)] = 2.540248 g H2O or 2.54 g H2O rounded to three significant figures
Answer: When 2.05 g of NH3 reacts with 3.76 g of O2, about 2.82 g of NO and 2.54 g of H2O form in the reaction.

Second question:
How many grams of the excess reactant remain after the limiting reactant is completely consumed?
First, we need to calculate how many grams of NH3 is needed to completely react with 3.76 g of O2. We can use dimensional analysis and equation coefficients to convert 3.76 g of O2 to moles of O2 to moles of NH3, to grams of NH3. We shall need the following equalities to set up conversion factors:
1 mol O2 = 31.9988 g O2
4 mol NH3 = 5 mol O2
1 mol NH3 = 17.03052 g NH3
[(3.76 g O2)/1][(1 mol O2)/(31.9988 g O2)][(4 mol NH3)/(5 mol O2)][(17.03052 g NH3)/(1 mol NH3)] = 1.6009284 g NH3
Since 1.6009282 g of NH3 are needed, and 2.05 g of NH3 are present, we can calculate the remaining grams by subtracting the required mass from the mass available:
(2.05 g)

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