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Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
=
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M
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.