Question

(45 pts) What is the theoretical yield (in g) of iron(III) carbonate that can be produced from 1.72 g of iron(III) nitrate and an excess of sodium carbonate? The molar mass of iron(III) carbonate (Fe2(CO3)3) is 291.73 g∙mol–1 and the molar mass of iron(III) nitrate (Fe(NO3)3) is 241.88 g∙mol–1.
2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

Answer 1

Answer:

1.04g of iron III carbonate

Explanation:

First, we must put down the equation of reaction because it must guide our work.

2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.

Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate

Mass of iron III nitrate reacted= 1.72g

Molar mass of iron III nitrate= 241.88 g∙mol–1

Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles

From the equation of the reaction;

2 moles of iron III nitrate yields 1 mole of iron III carbonate

7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate

Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass

Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate

Answer 2

Answer:

1.04g

Explanation:

Step 1:

The balanced equation for the reaction.

2Fe(NO3)3(aq) + 3Na2CO3(aq) → Fe2(CO3)3(s) + 6NaNO3(aq)

Step 2:

Determination of mass of Fe(NO3)3 that reacted and the mass of Fe2(CO3)3 that is produced from the balanced equation.

This is shown below:

Molar Mass of Fe(NO3)3 = 241.88g/mol

Mass of Fe(NO3)3 from the balanced equation = 2 x 241.88 = 483.76g

Molar Mass of Fe2(CO3)3 = 291.73g/mol

Summary:

From the balanced equation above,

483.76g of Fe(NO3)3 reacted to produce 291.73g of Fe2(CO3)3.

Step 3:

Determination of the theoretical yield of Fe2(CO3)3. This is illustrated below:

From the balanced equation above,

483.76g of Fe(NO3)3 reacted to produce 291.73g of Fe2(CO3)3.

Therefore, 1.72g of Fe(NO3)3 will react to produce = (1.72x291.73)/483.76 = 1.04g of Fe2(CO3)3

Therefore, the theoretical yield of Fe2(CO3)3 is 1.04g