Respuesta :
Answer:
Amount of CO required is 157.5 g
Explanation:
Molecular mass of Fe = 55.845 g/mol
Amount of Fe = 209.7 g
[tex]Mol\ of\ Fe=\frac{209.7}{55.845} \\=3.75\ mol[/tex]
Balanced reaction of reduction of Fe2O3 is as follows:
[tex]Fe_2O_3(s) + 3CO (g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]
From the balanced reaction, 2 mol of Fe is produced by 3 mol of CO.
Therefore, 3.75 mol of Fe will be produced by,
[tex]\frac{3}{2} \times 3.75 = 5.625\ mol\ CO[/tex]
Relation between mass and mol is as follows:
Mass = Mol × Molecular formula
Molecular mass of CO = 28 g/mol
Grams of CO required = 5.625 mol × 28 g/mol
= 157.5 g
Amount of CO required is 157.5 g.
Answer:
There are 157.8 grams of CO needed.
Explanation:
Step 1: Data given
Mass of Fe produced = 209.7 grams
Molar mass of Fe2O3 = 159.69 g/mol
Molar mass of Fe = 55.845 g/mol
Step 2: The balanced equation
3CO + Fe2O3 → 2Fe + 3CO2
Step 3: Calculate Moles of Fe
Moles Fe = mass Fe / molar mass Fe
Moles Fe = 209.7 grams / 55.845 g/mol
Moles Fe = 3.755 moles
Step 4: Calculate moles of CO
For 3 moles of CO we need 1 mol of Fe2O3 to produce 2 moles Fe and 3 moles of CO2
For 3.755 moles of Fe we need 3.755 *3/2 = 5.6325 moles of CO
Step 5: Calculate mass of CO
Mass CO = moles CO * molar mass CO
Mass CO = 5.6325 * 28.01 g/mol
Mass CO = 157.8 grams
There are 157.8 grams of CO needed.
