Question

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

Answer 1

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

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

0,2mol            :      0,25mol

Answer 2

Answer:

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

Explanation:

That's what i think it is