(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)

The theoretical yield of Fe₂(CO₃)₃ obtained from the reaction is 1.04 g

We'll begin by calculating the mass of Fe(NO₃)₃ that reacted and the mass of Fe₂(CO₃)₃ produced from the balanced equation.

2Fe(NO₃)₃ + 3Na₂CO₃ —> Fe₂(CO₃)₃ + 6NaNO₃

Molar mass of Fe(NO₃)₃ = 241.88 g/mol

Mass of Fe(NO₃)₃ from the balanced equation = 2 × 241.88 = 483.4 g

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

Mass of Fe₂(CO₃)₃ from the balanced equation = 1 × 291.73 = 291.73 g

From the balanced equation above,

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

From the balanced equation above,

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

Therefore,

1.72 g of Fe(NO₃)₃ will react to produce = (1.72 × 291.73) / 483.4 = 1.04 g of Fe₂(CO₃)₃

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

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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

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

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