Respuesta :
- Iron is in limited amount, limiting reagent
- The theoretical yield of iron(III) oxide formed is 27.1 grams.
Given:
The reaction between iron and dioxygen at high temperatures to form iron(III) oxide.
[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]
To find:
- The limiting reagent
- The theoretical yield of iron(III) oxide formed when Suppose 19.1 g of iron is reacted with 17.9 g of oxygen.
Solution:
- The mass of iron = 19.1 g
- The moles of iron [tex]=\frac{19.1 g}{55.845 g/mol}=0.342 mol[/tex]
- The mass of dioxygen = 19.1 g
- The moles of dioxygen [tex]=\frac{17.9 g}{15.999g/mol}=1.12 mol[/tex]
[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]
According to reaction, 4 moles of iron reacts with 3 moles of dioxygen then 0.342 moles of iron will react with:
[tex]=\frac{3}{4}\times 0.342 mol=0.2565 mol[/tex]
0.2465 moles of dioxygen will react with 0.342 moles of iron which means that:
- Iron is in limited amount, limiting reagent
- dioxygen is in an excessive amount, excessive reagent.
- The amount of iron(III) oxide formed will depend upon moles of iron.
[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex]
According to reaction, 4 moles of iron gives 2 moles of iron (III) oxide, then 0.342 moles of iron will give:
[tex]=\frac{2}{4}\times 0.342 mol= 0.171 \text{mol of }Fe_2O_3[/tex]
The theoretical yield of 0.171 moles of iron(III) oxide:
[tex]=0.171 mol\times 159.69 g/mol=27.1 g[/tex]
The theoretical yield of iron(III) oxide formed is 27.1 grams.
Learn more about limiting and excessive reagent here:
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