Question

H₂ (2) → H. H₂(8) + 2(g) wer

Answer 1

The given question is incomplete. The complete question is:

In the chemical reaction: , with 8 grams of and 16 grams of and the reaction goes to completion, what is the excess reactant and how much of that would remain?

A) 6 grams of  

B) 7 grams of  

C) 8 grams of

D) 12 grams of

E) 14 grams of

Answer: A) 6 grams of $H_2$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} H_2=\frac{8g}{2g/mol}=4moles$

$\text{Moles of} O_2=\frac{16g}{32g/mol}=0.5moles$

$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$  

According to stoichiometry :

1 moles of $O_2$ require 2 moles of $H_2$

Thus 0.5 moles of $O_2$ will require=$\frac{2}{1}\times 0.5=1.0moles$  of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

(4.0-1.0) = 3.0 moles of  are left unreacted

Mass of remained=

Thus 6.0 g of $H_2$ will remain.