Type the correct answer in the box. Express your answer to three significant figures. Iron(II) chloride and sodium carbonate rea
ct 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.
To solve this problem we have to find the moles of iron(II) chloride that react. Using the chemical equation, we can kknow moles FeCl2 = Moles FeCO3. Thus, we can find the moles of FeCO3. Converting these moles to grams using its molar mass -Molar mass FeCO3: 115.854g/mol-
<em>Moles FeCl2 = Moles FeCO3:</em>
1.24L * (2.00mol / L) = 2.48 moles FeCl2
<em>Mass FeCO3:</em>
2.48mol * (115.854g / mol) =
<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>