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fenix001 [56]
3 years ago
13

H₂ (2) → H. H₂(8) + 2(g) wer

Chemistry
1 answer:
Gre4nikov [31]3 years ago
4 0

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.

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