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.

Explanation:

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-

Moles FeCl2 = Moles FeCO3:

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

Mass FeCO3:

2.48mol * (115.854g / mol) =

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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