Note
The molar mass of N2O5
Element | Symbol | Atomic Mass | # of Atoms |
Nitrogen | N | 14.0067 | 2 |
Oxygen | O | 15.9994 | 5 |
2*14.0067+5*15.9994=108.01g/mol
The molar mass of NO2
Element | Symbol | Atomic Mass | # of Atoms |
Nitrogen | N | 14.0067 | 1 |
Oxygen | O | 15.9994 | 2 |
1*14.0067+2*15.9994=46.0055 g/mol
Number of moles= Given mass of the substance/molar mass of the substance
Answer
The reaction is,
2FeCl3 +3 K2CO3 Fe2 (CO3)3 + 6KCl.............................Balanced Chemical equation
which means,
Two moles of FeCl3 react with three moles of K2CO3 will produce one mole Fe2 (CO3)3 and six moles of KCl
or
For the complete reaction of 2 moles of FeCl3, 3 moles of K2CO3 required
or
For the complete reaction of 2/2 moles of FeCl3, 3/2 moles of K2CO3 required
OR
For the complete reaction of 1 mole of FeCl3, 1.5 moles of K2CO3 required
Here we have
65g of FeCl3 & 65g of K2CO3
Number of moles of FeCl3 6 5g of FeCl3 =Given mass FeCl3/ Molar mass of FeCl3
=65g /162.21g
=0.40 moles
Number of moles of K2CO3 6 5g of K2CO3 =Given mass K2CO3/ Molar mass of K2CO3
=65g/138.21g
=0.47Moles
Here the reaction is between,
0.40 moles FeCl3 & 0.47Moles of K2CO3
But For the complete reaction of 1 mole of FeCl3 , 1.5 moles of K2CO3 required
Hence
For the complete reaction of 10.40 mole of FeCl3 , 1.50.40 moles of K2CO3 required
OR
For the complete reaction of 10.40 mole of FeCl3 , 0.60 moles of K2CO3 required
But we have only 0.47Moles of K2CO3 , in this reaction amount of K2CO3 not sufficent for the complete reaction of 0.40 mole of FeCl3.
At this kind of situation, K2CO3 known as liminiting reagent.
In this reaction only K2CO3 is completely consumed, so we can calculate the amount of Fe2 (CO3)3 produced based on the amount of K2CO3 is consumed.
We know that,
Two moles of FeCl3 react with three moles of K2CO3 will produce one mole Fe2 (CO3)3 and six moles of KCl
OR
3 moles of K2CO3 will produce one 1 Fe2 (CO3)3
OR
1 moles of K2CO3 will produce one 1/3 moles Fe2 (CO3)3
Hence,
(10.47) Moles of K2CO3 will produce one (1/3 0.47) moles Fe2 (CO3)3
Number of moles ofFe2 (CO3)3 produced in this reaction=0.156 moles
The amount of Fe2 (CO3)3 in grams= Molar mass of Fe2 (CO3)3 Number of moles
=291.73 g/mol 0.156moes=45.70g
The amount of Fe2 (CO3)3 in grams=45.70g
Answer.2........................................................................................................................................................................................
The reaction is,
2N2O5 4NO2+ O2
2 Moles of N2O5dissociate and form 4 moles NO2
Or
1 Moles of N2O5dissociate and form 2 moles NO2
Here we have 721g of N2O5
Number of moles of N2O5 in 721g= Given mass of N2O5 / Molar mass of N2O5
The molar mass of N2O5 = 108.01 g/mol
=721g/108.1
=6.67 moles
We know that,
1 Moles of N2O5dissociate and form 2 moles NO2
Hence,
(1 6.67) Moles of N2O5dissociate and form (26.67) moles NO2
The amount ofNO2 formed in this reation=(26.67)=13.34 moles
The amount ofNO2 formed in this reaction in grams= Number of mole Molar mass of NO2
=13.34 moles46 g/mol
The amount ofNO2 formed in this reaction in grams =613.61g
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