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)
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).
To find the rate of loss of HBr gas, we need to determine the stoichiometry of the reaction, specifically the coefficients of the reactants and products. Once we have the stoichiometry, we can relate the rate of consumption of butane to the rate of loss of HBr gas.
Let's assume the balanced chemical equation for the reaction is:
C4H10 + 2HBr → C4H9Br + H2
From the balanced equation, we can see that for every 1 mole of butane (C4H10) consumed, 1 mole of HBr is also consumed. This means the rate of loss of HBr gas will be the same as 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 will also be 0.0333 mol/(L●s).
Therefore, the correct answer is c. 0.0333 mol/(L●s).