Question

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

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



_ mole(s) of iron (III) carbonate [Fe2CO3)3] is/are produced in the balanced equation.


a
4
b
1
c
3
d
2

Answer 1

0.616 moles of Fe₂(CO₃)₃ are produced when 200 g of FeCl₃ react

b. 1.

Step-by-step explanation:

The balanced reaction is:

• 2FeCl₃ + 3(NH₄)₂CO₃ → 6NH₄Cl + Fe₂(CO₃)₃

First we convert 200 grams of FeCl₃ into moles, using its molar mass:

• 200 g ÷ 162.2 g/mol = 1.23 mol FeCl₃

Then we convert 1.23 moles of FeCl₃ into moles of Fe₂(CO₃)₃, using the stoichiometric coefficients of the balanced reaction:

• 1.23 mol FeCl₃ *
  `(1molFe_2(CO_3)_3)/(2molFeCl_3)` = 0.616 mol Fe₂(CO₃)₃

The closest answer would be option b. 1.

Answer 2

Step-by-step explanation:

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

first balance the chemical equation

`2FeCl₃ + 3(NH₄)₂CO₃ → 6NH₄Cl + Fe₂(CO₃)₃`

2 mole. 3 mole. 6mole. 1mole

2*162g of FeCl₃ produce 236 g of Fe₂(CO₃)₃

200g of of FeCl₃

produce 236/(2*162)*20=145.68 g of Fe₂(CO₃)₃

1 mole of Fe₂(CO₃)₃=236g

145.68g of Fe₂(CO₃)=1/236*145.68=}0.61mole

closest answer is b 1

b.1mole(s) of iron (III) carbonate [Fe2CO3)3] is/are produced in the balanced equation