Question

For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(g) --> nitrogen monoxide(g)
a.What is the maximum mass of nitrogen monoxide that can be formed?

b. What is the FORMULA for the limiting reagent?

c. What mass of the excess reagent remains after the reaction is complete

Answer 1

Answer:

1. 10.8 g of NO

2. N₂ is the limting reagent

3. 3.2 g of O₂ does not react

Explanation:

We determine the reaction: N₂(g) + O₂(g) →  2NO(g)

We need to determine the limiting reactant, but first we need the moles of each:

5.04 g / 29 g/mol = 0.180 moles N₂

8.98 g / 32 g/mol = 0.280 moles O₂

Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂

Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO

Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 =  0.360 moles NO

If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g

As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.

As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.

0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.

Answer 2

Answer:

(a) 10.8 grams of nitrogen monoxide

(b) The formula for the limiting reagent is N

(c) 3.22 grams of the excess reagent remains

Explanation:

(a) The equation of reaction involves the reaction of 2 moles of nitrogen atom (N) and 1 mole of oxygen molecule (O2) to form 2 moles of nitrogen monoxide (NO).

From the equation of reaction above,

2 moles of nitrogen atom (28 grams) produced 2 moles of nitrogen monoxide (60 grams)

5.04 grams of nitrogen would produce (5.04×60)/28 = 10.8 grams of nitrogen monoxide.

(b) 1 mole of oxygen (32 grams) reacted with 2 moles of nitrogen (28 grams)

8.98 grams of oxygen would be required to react with (28×8.98)/32 = 7.86 grams of nitrogen but only 5.04 grams is available. Therefore, the limiting reagent is nitrogen atom and its formula is N.

(c) 2 moles of nitrogen monoxide (60 g) are produced from 1 mole of oxygen (32 g)

10.8 g of nitrogen monoxide is produced from (10.8×32)/60 = 5.76 grams of oxygen but 8.98 grams of oxygen is available. Therefore, oxygen is the excess reagent.

Mass of excess reagent that remains after the reaction is complete = 8.98 - 5.76 = 3.22 grams