In the following reaction, butane is consumed at the rate of 0.0333 mol/(L●s). What is the

rate of loss of HBr gas?
a. 0.0188 mol/(L●s)
b. 0.0165 mol/(L●s)
c. 0.0333 mol/(L●s)
d. 0.0667 mol/(L●s)

3 answers

To determine the rate of loss of HBr gas, we need to first understand the balanced chemical equation for the reaction.

The reaction between butane and HBr can be represented by the equation:

C4H10 + 2 HBr -> C4H9Br + H2

From the balanced equation, we can see that 2 moles of HBr gas are consumed for every 1 mole of butane consumed.

Therefore, the rate of loss of HBr gas is half of the rate of consumption of butane.

Given that the rate of consumption of butane is 0.0333 mol/(L*s), the rate of loss of HBr gas would be half of that.

Rate of loss of HBr gas = 0.0333 mol/(L*s) / 2 = 0.0165 mol/(L*s)

Therefore, the correct answer is b. 0.0165 mol/(L●s)
what is the correct answer?
I apologize for the confusion. The correct answer is indeed b. 0.0165 mol/(L●s).
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