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H(2) +Cl(2) rarr 2HCl" "Delta H =-184 kJ...

`H_(2) +Cl_(2) rarr 2HCl" "Delta H =-184 kJ`
`Delta H" for "36.5 xx 10^(-2) kg` is

A

`-92 kJ `

B

- 920 kJ

C

- 9200 kJ

D

- 9.2 kJ

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta H` for given balanced equation is for 2 moles =-184 kJ
`therefore" For "36.5xx10^(-2) kg HCl, Delta H" is "-920 kJ`
`Delta H =(36.5 xx10^(-2))/(36.5xx10^(-3))`
`=" 10 moles of HCl "=(-184xx10)/(2)=-920 kJ`
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