Type the correct answer in the box. Express your answer to three significant figures. 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? The reaction can produce grams of iron(II) carbonate.

The reaction can produce 287 grams of iron(II) carbonate

Explanation:

To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-

Moles FeCl2:

1.24L * (2.00mol / L) = 2.48 moles FeCl2

As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles

Mass FeCO3:

2.48mol * (115.854g / mol) =

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )