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lilavasa [31]
3 years ago
11

Need answer now! 100 points

Chemistry
2 answers:
ratelena [41]3 years ago
5 0

Answer: D

divide the moles of the substances given in the question by the co-efficient in the equation. Smallest quotient is the limiting reactant

s344n2d4d5 [400]3 years ago
3 0

Answer:

D) HBr

Explanation:

Given data:

Number of moles of Al = 8 mol

Number of moles of HBr = 8 mol

Limiting reactant = ?

Solution:

2Al + 6HBr      →      2AlBr₃ + 3H₂

now we will compare the moles of reactants with products.

                Al          :         AlBr₃

                   2         :            2

                    8         :          8

                 Al           :          H₂

                 2             :          3

                 8             :           3/2×8 = 12

                 HBr         :         AlBr₃

                   6           :              2

                    8           :            2/6×8 = 2.67

                 HBr         :            H₂

                   6            :            3

                   8             :           3/6×8 = 4

HBr produced less number of moles of product thus it will act as limiting reactant.

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BaLLatris [955]

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.

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kiruha [24]

Answer:

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Explanation:

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Take the HCF of carbon and hydrogen atoms :

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Then, we can write the formula as :

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