Respuesta :
This is an incomplete question, here is a complete question.
The balanced chemical reaction is:
[tex]6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2[/tex]
In another experiment, if 80 [tex]XO_3[/tex] molecules react with 104 [tex]BrZ_3[/tex] molecules. How many [tex]Br_2[/tex] molecules will be produced which reactant will be used up in the reaction.
Answer : The number of molecules of [tex]Br_2[/tex] will be, 52 molecules and [tex]BrZ_3[/tex] 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:
[tex]6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2[/tex]
First we have to determine the limiting reagent.
From the balanced reaction we conclude that,
As, 8 molecules of [tex]BrZ_3[/tex] react with 6 molecule of [tex]XO_3[/tex]
So, 104 molecules of [tex]BrZ_3[/tex] react with [tex]\frac{104}{8}\times 6=78[/tex] molecule of [tex]XO_3[/tex]
From this we conclude that, [tex]XO_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]BrZ_3[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the molecules of [tex]Br_2[/tex]
From the reaction, we conclude that
As, 8 molecules of [tex]BrZ_3[/tex] react to give 4 molecules of [tex]Br_2[/tex]
So, 104 molecules of [tex]BrZ_3[/tex] react to give [tex]\frac{104}{8}\times 4=52[/tex] molecules of [tex]Br_2[/tex]
Hence, the number of molecules of [tex]Br_2[/tex] will be, 52 molecules and [tex]BrZ_3[/tex] reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.
- The number of molecules of [tex]Br_{2}[/tex] will be, 52 molecules .
- The [tex]BrZ_{3}[/tex] reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.
The Chemical reaction is given as,
[tex]6XO_{3}+8BrZ_{3} [/tex] ⇒ [tex]6XZ_{4}+4Br_{2}+9O_{2}[/tex]
From above reaction, It is observed that, 8 molecules of [tex] BrZ_{3}[/tex] reacts with 6 molecules of [tex]XO_{3}[/tex].
So, 104 molecules of [tex]BrZ_{3}[/tex] reacts with,
[tex]=\frac{6}{8}*104=78 [/tex] molecules of [tex]XO_{3}[/tex]
Therefore, [tex] BrZ_{3}[/tex] is a limiting reagent.
From chemical reaction it is observed that,
8 molecules of [tex] BrZ_{3}[/tex] = 4 molecules of [tex]Br_{2}[/tex]
So, 104 molecules of [tex] BrZ_{3}[/tex] = [tex]\frac{4}{8}*108=52 [/tex] molecules of [tex]Br_{2}[/tex]
Hence, The number of molecules of [tex]Br_{2}[/tex] will be, 52 molecules .
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