The end product will depend upon
a) the amount of the reagent taken
b) the final treatment of the reaction
If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen
Figure 1
However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product
Figure 2
<u>Answer:</u> The mass of iron produced will be 77.6 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of FeO = 125 g
Molar mass of FeO = 71.8 g/mol
Putting values in equation 1, we get:
Given mass of aluminium = 25.0 g
Molar mass of aluminium = 27 g/mol
Putting values in equation 1, we get:
The given chemical reaction follows:
By Stoichiometry of the reaction:
2 moles of aluminium metal reacts with 3 mole of FeO
So, 0.93 moles of aluminium metal will react with = of FeO
As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.
Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of aluminium metal produces 3 mole of iron metal
So, 0.93 moles of aluminium metal will produce = of iron metal
Molar mass of iron = 55.85 g/mol
Moles of iron = 1.395 moles
Putting values in equation 1, we get:
Hence, the mass of iron produced will be 77.6 grams
