Question

For the following reaction,  10.1 grams of  nitrogen gas are allowed to react with  5.97 grams of  oxygen gas .

nitrogen(g) +  oxygen(g)   nitrogen monoxide(g)

What is the maximum mass of  nitrogen monoxide that can be formed?   grams

What is the  FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?   grams

Answer 1

1_

Molar mass of N2 = 28.02 g/mol


mass(N2)= 10.1 g

use:
number of mol of N2,
n = mass of N2/molar mass of N2
=(10.1 g)/(28.02 g/mol)
= 0.3605 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 5.97 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(5.97 g)/(32 g/mol)
= 0.1866 mol
Balanced chemical equation is:
N2 + O2 ---> 2 NO


1 mol of N2 reacts with 1 mol of O2
for 0.3605 mol of N2, 0.3605 mol of O2 is required
But we have 0.1866 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of NO,
MM = 1*MM(N) + 1*MM(O)
= 1*14.01 + 1*16.0
= 30.01 g/mol

According to balanced equation
mol of NO formed = (2/1)* moles of O2
= (2/1)*0.1866
= 0.3731 mol


use:
mass of NO = number of mol * molar mass
= 0.3731*30.01
= 11.2 g
Answer: 11.2 g

2)
Answer: O2

3)
According to balanced equation
mol of N2 reacted = (1/1)* moles of O2
= (1/1)*0.1866
= 0.1866 mol
mol of N2 remaining = mol initially present - mol reacted
mol of N2 remaining = 0.3605 - 0.1866
mol of N2 remaining = 0.1739 mol


Molar mass of N2 = 28.02 g/mol

use:
mass of N2,
m = number of mol * molar mass
= 0.1739 mol * 28.02 g/mol
= 4.873 g
Answer: 4.87 g