WCl6 + H2 + CH4 WC + 6HCl
If 23.5 g WCl6 reacts with excess H2 and CH4, and 10.8 g WC are recovered, what is the percent yield?
Taking W=183.84 Cl=35.453 C=12.0107
WCl6= 183.84+ (35.453×6)= 396.558g
WC= 183.84+ 12.0107
= 195.8507g
396.558 of WCl6 yields 195.8507 of WCl6
23.5g of WCl6 in the same proportion should yield
"=\\frac{23.5}{396.558}\u00d7195.8507=11.6061g"
of WCl6
Actual yield given = 10.8g
"\\%" yield "=\\frac{Actual \\>yield}{Theoretical \\> value}\u00d7100\\%"
"=\\frac{10.8}{11.6061}\u00d7100\\%=93.06\\%"
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