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When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to ch4(g) + cl2(g) → ch3cl(g) + hcl(g)?
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When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to ch4(g) + cl2(g) → ch3cl(g) + hcl(g)?

User BuddhistBeast
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1 Answer

1 vote

Answer:

the answer is 57.03%

Step-by-step explanation:

%yield= ((practical yield)/(theoretical yield))×100%

When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl-example-1
User Steve Miskovetz
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