Iron(II) chloride and sodium carbonate react to make iron(II) carbonate and sodium chloride: FeCl2(aq) + Na2CO3(s) → FeCO3(s) + 2NaCl(aq).

Given 1.24 liters of a 2.00 M solution of iron(II) chloride and unlimited sodium carbonate,
how many grams of iron(II) carbonate can the reaction produce?

287.68g

Explanation:

We'll begin by calculating the number of mole of FeCl2 in 1.24L of a 2M FeCl2 solution. This is illustrated below:

Molarity = 2M

Volume = 1.24L

Mole of FeCl2 =..?

Mole = Molarity x Volume

Mole of FeCl2 = 2 x 1.24

Mole of FeCl2 = 2.48 moles

Next, the balanced equation for the reaction.

FeCl2(aq) + Na2CO3(s) → FeCO3(s) + 2NaCl(aq).

Next, we shall determine the mole of FeCO3 produced from the reaction.

This is illustrated below:

From the balanced equation above,

1 mole of FeCl2 reacted to produce 1 mole of FeCO3.

Therefore, 2.48 moles of FeCl2 will also react to produce 2.48 moles of FeCO3.

Finally, we shall convert 2.48 moles of FeCO3 to grams.

Number of mole of FeCO3 = 2.48 moles

Molar mass of FeCO3 = 56 + 12 + (16x3) = 116g/mol.

Mass of FeCO3 =..?

Mass = mole x molar mass

Mass of FeCO3 = 2.48 x 116

Mass of FeCO3 = 287.68g

Therefore, 287.68g of FeCO3 were from the reaction.