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H₂ (2)
→ H.
H₂(8)
+
2(g)
wer


Sagot :

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 [tex]H_2[/tex]

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

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

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

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]  

According to stoichiometry :

1 moles of [tex]O_2[/tex] require 2 moles of [tex]H_2[/tex]

Thus 0.5 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.5=1.0moles[/tex]  of [tex]H_2[/tex]

Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.

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

Mass of remained=

Thus 6.0 g of [tex]H_2[/tex] will remain.