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
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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