Answer :
Answer:
6.37g of iron are produced.
Explanation:
1st) From the balanced reaction we know that 1 mole of Fe2O3 reacts with 3 moles of H2 to produce 2 moles of Fe and 3 moles of H2O.
With the molar mass of Fe2O3 (160g/mol) and H2 (2g/mol) we can convert the moles to grams:
- Fe2O3 conversion: 1 mole is equal to 160g, so here it is not necessary to calculate the conversion.
- H2 conversion:
[tex]3moles*\frac{2g}{1mole}=6g[/tex]So, from the balanced reaction we know that 160g of Fe2O3 react with 6g of H2.
2nd) Now we have to find out which compound is the limiting reactant and which compound is the excess reactant, using the 9.1g and 16.8g given in the exercise:
[tex]\begin{gathered} 160gFe_2O_3-6gH_2 \\ 9.1gFe_2O_3-x=\frac{9.1gFe_2O_3*6gH_2}{160gFe_2O_3} \\ x=0.34gH_2 \end{gathered}[/tex]So, the 9.1g of Fe2O3 will need 0.34g of H2 to react properly, but we have 16.8g of H2 so H2 is the excess reactant and Fe2O3 is the limiting reactant.
3rd) Using the limiting reactant, we can calculate the grams of iron that are produced.
It is necessary to use the molar mass of iron (56g/mol) to convert the 2 moles in the reaction to grams (2 Fe moles = 112g).
[tex]\begin{gathered} 160gFe_2O_3-112gFe \\ 9.1gFe_2O_3-x=\frac{9.1gFe_2O_3*112gFe}{160gFe_2O_3} \\ x=6.37gFe \end{gathered}[/tex]Finally, 6.37g of iron are produced.
