Question

In a lab experiment, 7.97 g of phosphorus reacts with bromine to form 69.65 g of phosphorus tribromide. How much phosphorus tribromide would be formed if 12.05 g of phosphorus reacted with 61.68 g of bromine? answer:______gi put what i got in the image and it is wrong sorry

Answer 1

1) List the known and unknown quantities.

First experiment:

Reactants

Phosphorus: 7.97 g.

Bromine: excess.

Product

Phosphorus tribromide: 69.65 g.

Second experiment

Reactants

Phosphorus: 12.05 g.

Bromine: 61.68 g.

Product

Phosphorus tribromide: unknown.

2) Write and balance the chemical equation.

`2P+3Br_2\rightarrow2PBr_3`

3) Convert the masses.

3.1-Convert the mass of P to moles of P.

The molar mass of P is 30.97 g/mol.

`mol\text{ }P=12.05\text{ }g\text{ }P*\frac{1\text{ }mol\text{ }P}{30.97\text{ }g\text{ }P}=0.38909\text{ }mol\text{ }P`

3.2-Convert the mass of Br to moles of Br.

The molar mass of Br2 is 159.8080 g/mol.

`mol\text{ }Br_2=61.68\text{ }g\text{ }Br_2*\frac{1\text{ }mol\text{ }Br_2}{159.8080\text{ }g\text{ }Br_2}=0.38596\text{ }mol\text{ }Br_2`

4) Which is the limiting reactant?

4.1-How many moles of Br2 do we need to use all of the P?

The molar ratio between Br2 and P is 3 mol Br2: 2 mol P.

`mol\text{ }Br_2=0.38909\text{ }mol\text{ }P*\frac{3\text{ }mol\text{ }Br_2}{2\text{ }mol\text{ }P}=0.583635\text{ }mol\text{ }Br_2`

We need 0.583635 mol Br2 and we have 0.38596 mol Br2. We do not have enough Br2. This is the limiting reactant.

4.2-How many moles of P do we need to use all of the Br2?

The molar ratio between Br2 and P is 3 mol Br2: 2 mol P.

`mol\text{ }P=0.38596\text{ }mol\text{ }Br_2*\frac{2\text{ }mol\text{ }P}{3\text{ }mol\text{ }Br_2}=0.25731\text{ }mol\text{ }P`

We need 0.25731 mol P and we have 0.38909 mol P. We have enough P. This is the excess reactant.

5) Moles of phosphorus tribromide produced from the limiting reactant.

Limiting reactant: 0.38596 mol Br2.

The molar ratio between Br2 and PBr3 is 3 mol Br2: 2 mol PBr2.

`mol\text{ }PBr_2=0.38596\text{ }mol\text{ }Br_2*\frac{2\text{ }mol\text{ }PBr_3}{3\text{ }mol\text{ }Br_2}=0.25731\text{ }mol\text{ }PBr_3`

6) Conver the moles of Pbr3 to mass of PBr3.

The molar mass of PBr3 is 270.6858 g/mol.

`g\text{ }PBr_3=0.25731\text{ }PBr_3*\frac{270.6858\text{ }g\text{ }PBr_3}{1\text{ }mol\text{ }PBr_3}=69.6502\text{ }g\text{ }PBr_3`

The mass of PBr3 produced in the reaction is 69.65 g PBr3.

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