Question

In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant will be used up in the reaction

Answer 1

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

$6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2$

In another experiment, if 80 $XO_3$ molecules react with 104 $BrZ_3$ molecules. How many $Br_2$ molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of $Br_2$  will be, 52 molecules and $BrZ_3$ reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

$6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2$

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of $BrZ_3$ react with 6 molecule of $XO_3$

So, 104 molecules of $BrZ_3$ react with $\frac{104}{8}\times 6=78$ molecule of $XO_3$

From this we conclude that, $XO_3$ is an excess reagent because the given moles are greater than the required moles and $BrZ_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of $Br_2$

From the reaction, we conclude that

As, 8 molecules of $BrZ_3$ react to give 4 molecules of $Br_2$

So, 104 molecules of $BrZ_3$ react to give $\frac{104}{8}\times 4=52$ molecules of $Br_2$

Hence, the number of molecules of $Br_2$  will be, 52 molecules and $BrZ_3$ reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.