Iron 3 Nitrate Molar Mass
(45 pts) What is the theoretical yield (in g) of fe(III) carbonate that can be produced from i.72 g of iron(3) nitrate and a
n excess of sodium carbonate? The molar mass of iron(3) carbonate (Fe2(CO3)3) is 291.73 1000∙mol–i and the tooth mass of atomic number 26(Iii) nitrate (Atomic number 26(NO3)3) is 241.88 g∙mol–1.
2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(southward) + 6NaNO3(aq)
2 answers:
Answer:
1.04g of iron III carbonate
Explanation:
First, nosotros must put down the equation of reaction considering it must guide our work.
2Fe(NO3)iii(aq) + 3Na2CO3(aq)→Fe2(CO3)3(southward) + 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 fe 3 nitrate reacted/ tooth mass of atomic number 26 Iii nitrate
Mass of iron III nitrate reacted= 1.72g
Molar mass of atomic number 26 Three nitrate= 241.88 g∙mol–1
Number of moles of iron III nitrate= 1.72g/241.88 one thousand∙mol–one= vii.xi×10^-3 moles
From the equation of the reaction;
two moles of iron Three nitrate yields 1 mole of iron Iii carbonate
7.xi×10^-3 moles moles of iron 3 nitrate yields 7.11×10^-iii × 1/ ii= 3.56×x^-3 moles of iron III carbonate
Theoretical mass yield of fe Three carbonate = number of moles of iron III carbonate × tooth mass
Theoretical mass yield of iron 3 carbonate = 3.56×ten^-3 moles ×291.73 g∙mol–i = i.04g of iron III carbonate
Answer:
1.04g
Caption:
Step ane:
The balanced equation for the reaction.
2Fe(NO3)iii(aq) + 3Na2CO3(aq) → Fe2(CO3)3(southward) + 6NaNO3(aq)
Stride two:
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 Iron(NO3)3 = 241.88g/mol
Mass of Fe(NO3)3 from the balanced equation = 2 x 241.88 = 483.76g
Tooth Mass of Fe2(CO3)3 = 291.73g/mol
Summary:
From the counterbalanced equation above,
483.76g of Iron(NO3)3 reacted to produce 291.73g of Fe2(CO3)3.
Step 3:
Determination of the theoretical yield of Fe2(CO3)iii. This is illustrated beneath:
From the balanced equation higher up,
483.76g of Fe(NO3)three reacted to produce 291.73g of Fe2(CO3)3.
Therefore, 1.72g of Iron(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
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Iron 3 Nitrate Molar Mass,
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