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oksian1 [2.3K]
3 years ago
8

How many moles of Ca3 (PO4)2 can be formed when 1.9 moles of CaCO3 react with 2.7 moles of FePO4?

Chemistry
1 answer:
Nonamiya [84]3 years ago
5 0

Answer:-

0.633 moles

Explanation:-

The first step to solving this problem is to write the balanced chemical equation.

The balanced chemical equation for this reaction is

3 CaCO3 + 2 FePO4 = Ca3(PO4)2 + Fe2(CO3)3

From the balanced chemical equation, we can see

3 moles of CaCO3 react with 2 moles of FePO4.

In the question we are told there are 1.9 moles of CaCO3.

So 1.9 moles of CaCO3 react with (2 moles x 1.9 moles )/ 3 moles= 1.267 moles of FePO4.

The question tells us there are 2.7 moles of FePO4.

So there is excess FePO4.

Thus our limiting reagent is CaCO3. It will determine how much Ca3(PO4)2 is formed.

From the balanced chemical equation, we can see

3 moles of CaCO3 gives 1 mole of Ca3(PO4)2

1.9 moles of CaCO3 gives (1 moles x 1.9 moles ) / 3 moles= 0.633 moles of Ca3(PO4)2.

0.633 moles of Ca3 (PO4)2 can be formed when 1.9 moles of CaCO3 react with 2.7 moles of FePO4

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1 L ------- 1000 cm³
1.45 L ----- ???

1.45 * 1000 = 1450 cm³  ( volume ) 

Density = 0.710 g/cm³

mass =  in Kg 

m = D * V

m = 0.710  * 1450

m = 1029.5 g

1 Kg ------- 1000 g
   kg -------- 1029.5 g

mass = 1029.5 / 1000

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hope this helps!

7 0
3 years ago
In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha
devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

<u>Step 1:</u> Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

<u>Step 2:</u> The balanced equation

N2 + 3H2 → 2NH3

<u>Step 3:</u> Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

<u>Step 4:</u> Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

<u>Step 5</u>: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

<u>Step 6: </u>Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

<u>Step 7:</u> The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0
4 years ago
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