Question

please help :)

4NH_3+ 〖3 O〗_2 → 2 NO+6 H_2 O

1. How many grams of NO can be produced from 12 g of NH3 and 12 g of O2?

2. What is the limiting reactant? What is the excess reactant?

3. How much excess reactant remains when the reaction is over?

Answer 1

$\\ \tt\hookrightarrow 4NH_3+3O_2\longrightarrow 2NO+6H_2O$

#1

We can solve through ammonia or oxygen .Lets go through ammonia

• 4mol of ammonia produces 2mol NO
• 2mol of ammonia produces 1mol NO
• 1mol of ammonia produces 0.5mol NO.

Moles of Ammonia

$\\ \tt\hookrightarrow \dfrac{12}{17}=0.7mol$

Moles of NO

$\\ \tt\hookrightarrow 0.7(0.5)=0.35mol$

Mass of NO

• 0.35(30)=10.5g

#2

NH_3 is excess reagent and O_2 is limiting reagent .

#3

We need ∆n

• ∆n=8-7=1mol