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Consider the following, balanced reaction: 4 NH 3 ( g ) + 5 O 2 ( g ) --> 4 NO(

ID: 1002574 • Letter: C

Question

Consider the following, balanced reaction:

4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(l)

Which of the following are TRUE? There may be more than one correct answer.

Select one or more:

The reaction shown has exactly 4 moles of NH3 and 5 moles of O2 present at the beginning of the reaction.

If 8 moles of NH3 are used up, then 10 moles of O2 are also used up.

If you have 4 moles of NH3 and 4 moles of O2, then O2 is the limiting reactant.

If you have 4 moles of NH3 and 4 moles of O2, then NH3 is the limiting reactant.

For every 1 mole NH3 used up, there is 1 mole NO created.

For every 5 grams of O2 that react, 6 grams of H2O are created.

Explanation / Answer

4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(l)

(i) The reaction shown has exactly 4 moles of NH3 and 5 moles of O2 present at the beginning of the reaction.

We can see in the balanced equation, that there are 4 mol of NH3, 5 mol of O2 in the reactant side. So, the statement is true.

(ii) If 8 moles of NH3 are used up, then 10 moles of O2 are also used up.

From equation, for 4 mol of NH3, 5 mol of O2 is used. If we double both the moles, then for 8 mol of NH3, 10 mol of O2 will be used. So, this statement is also true.

(iii) If you have 4 moles of NH3 and 4 moles of O2, then O2 is the limiting reactant.

From equation, for every 4 mol of NH3, 5 mol of O2 is used. But if 4 mol of O2 is there, then it means that O2 is limiting reactant and NH3 is in excess. So, this statement is also true.

(iv) If you have 4 moles of NH3 and 4 moles of O2, then NH3 is the limiting reactant.

From equation, for every 4 mol of NH3, 5 mol of O2 is used. But if 4 mol of O2 is there, then it means that O2 is limiting reactant and NH3 is in excess. So, this statement is also false.

(v) For every 1 mole NH3 used up, there is 1 mole NO created.

From equation, 4 mol of NH3 produces 4 mol of NO. So, 1 mol of NH3 will produce 1 mol of NO. This statement is true.

(vi) For every 5 grams of O2 that react, 6 grams of H2O are created.

From equation, 5 mol of O2 produces 6 mol of H2O.

Number of moles in 5 grams of O2 = Mass/molar mass of O2 = 5/32 = 0.156 mol

Number of moles in 6 grams of H2O = Mass/molar mass of H2O = 6/18 = 0.33 mol

If 5 mol of O2 produces 6 mol of H2O.

Then, 0.156 mol of O2 will produce H2O = (6/5)*0.156 = 0.187 mol

But given moles of H2O produced = 0.33

So, this statement is false.