Answer:
4.13 moles of Fe will be produced in the reaction
Explanation:
To produce Fe from CO, we need to react Fe₂O₃ and the carbon monoxide.
The reaction is:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
1 mol of iron (III) oxide react to 3 moles of carbon monoxide in order to produce 2 moles of Iron and 3 moles of carbon dioxide.
We assume, the Fe₂O₃ in excess, so we propose this rule of three
3 moles of CO can produce 2 moles of Fe
Therefore 6.20 moles of CO will produce (6.20 . 2)/3 = 4.13 moles of Fe
Answer : The number of moles of sulfur needed to oxidize will be, 3 moles
Solution : Given,
Moles of zinc = 3 moles
The balanced reaction will be,
By the stoichiometry, 1 mole of 
ion react with the 1 mole of 
to give 1 mole of zinc sulfide.
From the balanced reaction, we conclude that
As, 1 mole of zinc react with 1 mole of sulfur
So, 3 moles if zinc react with 3 moles of sulfur
Hence, the number of moles of sulfur needed to oxidize will be, 3 moles
Looks like it would be A................
Answer:
fraction of vacancies for this metal FV = 1.918*10⁻⁴
Explanation:
Given:
The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )
But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol
Density of the material is given in g/cm3 we need to convert it to g/m³
Density of material ( p ) in g/m³ :
To convert we know that
1 g/cm³ = 1000000 g/m³ then
7.40 * ( 1000000 ) = 7.40*10⁶ g/m³
So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³
Given Atomic mass = 85.5 g/mol
To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above
N = NA * p / A
N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5
N = 4.45*10³⁰ / 85.5
N = 5.212*10²⁸ atoms/m³
We can now Calculate the fraction of vacancies using the formula below;
Fv = Nv / N
Fv = 1*10²⁵ / 5.212*10²⁸
fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴
1: Decomposition reaction
2: Combination reaction
3: product
4: Reactant
