All of the following methods would increase the solubility of a solid solute, except for
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14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.
Explanation:
The balanced equation for the reaction is to be known so that number of moles taking part can be known.
The balanced chemical equation is
4Fe + 3⇒ 2
From the given weight of iron to be used for the production of , number of moles of Fe taking part in the reaction can be known by the formula:
Number of moles= mass ÷ Atomic mass of one mole of the element.
(Atomic weight of Fe is 55.845 gm/mole)
Putting the values in equation
Number of moles = 10 gm ÷ 55.845 gm/mole
= 0.179 moles
Applying the stoichiometry concept
4 moles of Fe gives 2 Moles of Fe2O3
0.179 moles will produce x moles of Fe2O3
So, 2÷ 4 = x ÷ 0.179
2/4 = x/ 0.179
2 × 0.179 = 4x
2 × 0.179 / 4 = x
x = 0.0895 moles
So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.
Now the formula used above will give the weight of Fe2O3
weight = atomic weight × number of moles
= 159.69 grams × 0.0895
= 14.292 grams of Fe2O3 formed.
Explanation:
Molarity is defined as number of moles per liter of solution.
Mathematically, molarity =
It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.
molarity =
0.0800 M =
no. of moles = 1.6 mol
Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.
No. of moles =
mass in grams =
=
= 399.488 g
Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.