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rusak2 [61]
3 years ago
7

Under the right conditions aluminum will react with chlorine to produce aluminium chloride.

Chemistry
1 answer:
Andreas93 [3]3 years ago
5 0

Answer:

Ok

The Limiting reagent is Aluminum. Its not in excess. So the reaction ends once its exhausted. So all Aluminum will be used up

From the question ....

157g of Al

Find the number if reacting Moles...

M=mass/molar mass

M= 157/27

=5.81moles of Aluminum is reacting.

Now looking at the equation of reaction(Always make sure its balanced)...

2moles of Al react to produce 2 Moles of AlCl3

Their Mole ratio is equal

Since their Mole ratio is equal 2:2.... That means... That 5.81moles of AlCl3 would be produced because the mole of Al is 5.81

So We now know the Moles of AlCL3 produced

getting the Mass won't be difficult.

Again

Mole=Mass/Mm

Mass = Mole x Mm(Molar Mass)

Mass of AlCl3 = 5.81 x (27 + 35.5x3)

Note: The arithmetic done in the parenthesis is the molar mass of AlCl3

Mass of AlCl3 = 5.81 x 133.5

=775.63g of AlCl3.

Have a great Day!!!!

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Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c
Anna [14]

The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

d)

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

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4 years ago
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