Question

What is the limiting reactant if 4.0 g of NH3 react with 8.0 g of oxygen?

Answer 1

A. O2 because it produces only 0.20 mol of NO.

Step-by-step explanation:

That's what i think it is

Answer 2

`NH_(3):\\\\ M_{NH_(3)}=17(g)/(mol)\\ m=4g \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ n=(m)/(M)=(4g)/(17(g)/(mol))\approx0,235mol\\\\\\\ O_(2):\\\\ M_{O_(2)}=32(g)/(mol)\\ m=8g \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ n=(m)/(M)=(8g)/(32(g)/(mol))=0,25mol`

4NH₃ + 5O₂ ⇒ 4NO + 6H₂O
4mol : 5mol
0,235mol : 0,25mol
limiting reagent

0,2mol : 0,25mol