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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.

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Answer:

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 reaction can produce 287 grams of iron(II) carbonate

Answer:

287

Explanation:

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