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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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Answer:

The temperature in degress Celsius is 52.25°C

Explanation:

According the equation:

V(\rho _{a}  -\rho _{t})g=mg\\\rho _{a}  -\rho _{t}=\frac{n}{V} =\frac{340}{2950} =0.115kg/m^{3}

\rho _{t}=\rho _{a}  -0.115=1.2-0.115=1.085kg/m^{3}

The temperature is:

T=\frac{P}{\rho _{t} R} =\frac{1.013x10^{5} }{1.085*287.05} =325.25K=52.25°C

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4 years ago
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Is it an atomic model? I'm not really sure
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Why would fluorine, chlorine, and bromine be placed in the same group (the halogens) in the periodic table of elements? (1 point
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4 years ago
I will give brainliest if done well.
topjm [15]

Answer:

8.37g

Explanation:

Step 1 :

The balanced equation for the reaction. This is given below:

N2 + 2O2 —> 2NO2

Step 2:

Data obtained from the question.

Volume (V) of N2 = 2L

Pressure (P) = 840mmHg

Temperature (T) = 24°C

Number of mole (n) of N2 =?

Step 3:

Conversion to appropriate unit.

For pressure:

760mmHg = 1atm

840mmHg = 840/760 = 1.11 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 24°C

T(K) = 24°C + 273

T(K) = 297K

Step 4:

Determination of the number of mole N2.

The number of mole of N2 can be obtained by using the ideal gas equation as follow:

Volume (V) of N2 = 2L

Pressure (P) = 1.11 atm

Temperature (T) = 297K

Number of mole (n) of N2 =?

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

Divide both side by RT

n = PV / RT

n = 1.11 x 2 / 0.082 x 297

n = 0.091 mole

Therefore, the number of mole of N2 that reacted is 0.091 mole

Step 5:

Determination of the mass of NO2 produced from the reaction. This is illustrated below:

N2 + 2O2 —> 2NO2

From the balanced equation above,

1 mole of N2 produced 2 moles of NO2.

Therefore, 0.091 mole of N2 will produce = 0.091 x 2 = 0.182 mole of NO2.

Finally, we will convert 0.182 mole of NO2 to gram as shown below:

Number of mole NO2 = 0.182 mole

Molar mass of NO2 = 14 + (16x2) = 46g/mol

Mass = number of mole x molar mass

Mass of NO2 = 0.182 x 46

Mass of NO2 = 8.37g

7 0
3 years ago
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evablogger [386]

Answer:

b

Explanation:

b

4 0
3 years ago
Read 2 more answers
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