Question

200 grams of iron (III) chloride reacts with ammonium carbonate [(NH4)2CO3] in the following equation.

FeCl3 + (NH4)2CO3 ---------> NH4Cl + Fe2(CO3)3



The number of grams of iron (III) carbonate Fe2(CO3)3 that will be produced is _ grams.



a
181
b
198
c
229
d
395

Answer 1

Answer. First things first... Balance the Chemical Equation

2FeCl3 + 3(NH4)2CO3 ---------> 6NH4Cl + Fe2(CO3)3

Its balanced Now.

200g of FeCl3 ---Given this... We can get jts reacting Moles

Mole = Mass/Molar Mass

Molar Mass of FeCl3 = 56 + (35.5)x3=162.5gmol-¹

Mole=200/162.5 =1.231moles of FeCl3

From the equation of Reaction

When 2mole of FeCl3 reacts.... It produces 1mole of Iron III carbonate [Fe2(CO3)3]

Since Mole ratio of reaction between both is 2:1

Therefore When 1.231moles of FeCl3 react....

1.231/2 moles of Fe2(CO3)3 will be produced.

= 1.231/2 = 0.616moles.

0.616Moles of Fe2(CO3)3 will be produced

Recall MASS =MOLE X MOLAR MASS

MOLAR MASS OF FE2(CO3)3= 56(2) + (60)(3)=292gmol-¹

MASS = 0.616 x 292 =179.872g ~ 180g of Fe2(CO3)3 Will Be Produced.

OPTION A SEEMS TO BE THE CLOSEST. THAT SHOULD BE OUR ANSWER. I MAY HAVE MISSED IT A BIT BECAUSE OF SOME APPROXIMATIONS.

BYE!

Answer 2

`FeCl₃ + (NH₄)₂CO₃ → NH₄Cl + Fe₂(CO₃)₃`

first balance the chemical equation

first balance the chemical equation
`2FeCl₃ + 3(NH₄)₂CO₃ → 6NH₄Cl + Fe₂(CO₃)₃`

2FeCl₃=2*162=324g

6NH₄Cl=6*53.5=321g

324 g of FeCl₃ gives 321 g of NH₄Cl

200g of FeCl₃ gives 321/324*200=198.14 g of NH₄Cl

in nearest value it is 198g/mole

The number of grams of ammonium chloride (NH4Cl) that will be produced is C.198___ grams.