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Tom [10]
3 years ago
11

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed

Chemistry
1 answer:
MaRussiya [10]3 years ago
3 0

The question is incomplete, here is the complete question:

Suppose the reaction is:

A+2B\rightarrow C

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed

<u>Answer:</u> The maximum amount of C produced will be 1 mole

<u>Explanation:</u>

We are given:

Moles of A = 6 moles

Moles of B = 2 moles

For the given chemical reaction:

A+2B\rightarrow C

By Stoichiometry of the reaction:

2 moles of B reacts with 1 mole of A

So, 2 moles of B will react with = \frac{1}{2}\times 2=1 mole of A

As, As, given amount of A is more than the required amount. So, it is considered as an excess reagent.

Thus, B is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of B produces 1 mole of C

So, 2 moles of given B will produce =\frac{1}{2}\times 2=1mol of C

Hence, the maximum amount of C produced will be 1 mole

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How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
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The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

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Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

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Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

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Step 3:

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The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

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Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

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