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3 years ago

Calculate the mass in grams of 8.35 × 1022 molecules of CBr4.

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

4 0

Answer: 45.983 g CBr₄

Explanation:

To convert from moles to grams, you know that we will need molar mass and Avogadro's number.

Avogadro's number: 6.022×10²³ molecules/mol

Molar mass: 331.627 grams/mol

Now that we have what we need, you can use these to solve for grams. $8.35*10^2^2molecules CBr_{4} *\frac{1mol}{6.022*10^2^3molecules} *\frac{331.627 g}{1 mol} =45.983 g$

Our final answer is 45.983 g CBr₄.

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When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g

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Answer: The mass of iron produced will be 77.6 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For FeO:

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol$

For aluminium:

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol$

The given chemical reaction follows:

$3FeO+2Al\rightarrow 3Fe+Al_2O_3$

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = $\frac{3}{2}\times 0.93=1.395mol$ of FeO

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

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

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = $\frac{3}{2}\times 0.93=1.395moles$ of iron metal

Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

$1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g$

Hence, the mass of iron produced will be 77.6 grams

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