Respuesta :
Answer :
(a) The number of grams of iron metal obtained, 698.616 g
(b) The number of grams of carbon dioxide used in the reaction, 826.32 g
Solution : Given,
Mass of hematite = 1 Kg = 1000 g
Molar mass of hematite = 159.69 g/mole
Molar mass of iron = 55.8 g/mole
Molar mass of carbon dioxide = 44 g/mole
(a) First we have to calculate the moles of hematite.
[tex]\text{Moles of hematite}=\frac{\text{Mass of hematite}}{\text{Molar mass of hematite}}=\frac{1000g}{159.69g/mole}=6.26moles[/tex]
Now we have to calculate the moles of iron.
The given balanced reaction is,
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]
From the balanced reaction, we conclude that
As, 1 mole of hematite react to give 2 moles of iron
So, 6.26 moles of hematite react to give [tex]2\times 6.26=12.52[/tex] moles of iron
Now we have to calculate the mass of iron.
[tex]\text{Mass of iron}=\text{Moles of iron}\times \text{Molar mass of iron}[/tex]
[tex]\text{Mass of iron}=(12.52moles)\times (55.8g/mole)=698.616g[/tex]
(b) Now we have to calculate the moles of carbon dioxide.
From the balanced reaction we conclude that
As, 1 mole of hematite react to give 3 moles of carbon dioxide
So, 6.26 moles of hematite react to give [tex]3\times 6.26=18.78[/tex] moles of carbon dioxide
Now we have to calculate the mass of carbon dioxide.
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]
[tex]\text{Mass of }CO_2=(18.78moles)\times (44g/mole)=826.32g[/tex]
Therefore, (a) The number of grams of iron metal obtained, 698.616 g
(b) The number of grams of carbon dioxide used in the reaction, 826.32 g