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bearhunter [10]
3 years ago
5

If you had excess aluminum, how many moles of aluminum chloride could be produced from 21.0 g of chlorine gas, cl2?

Chemistry
1 answer:
AleksandrR [38]3 years ago
3 0
When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:

2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:

(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃

Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
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Type the correct answer in the box. Express your answer to three significant figures.
VladimirAG [237]

<u>Given:</u>

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

<u>To determine:</u>

Theoretical yield of calcium phosphate, Ca3(PO4)2

<u>Explanation:</u>

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 =  96.1 g

Molar mass of Ca(NO3)2 =  164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles

Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g



7 0
3 years ago
Read 2 more answers
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4vir4ik [10]
Matter is anything that has mass and occupies space, It can exist in 3 states, or phases: solid, liquid, and gas.

Solid molecules are closely packed together and retain a fixed shape.

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Gas molecules are far apart and fill the container

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7 0
2 years ago
Read 2 more answers
What volume of 0.194 MNa3PO4 solution is necessary to completely react with 85.5 mL of 0.109 MCuCl2 ?
ira [324]

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴  1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

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Answer:

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6 0
2 years ago
According to rounding rules for addition the sum of 27.1, 34.538, and 37.68 is
Katena32 [7]
The answer should be...99.318!
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