Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
The reaction is
FeO + Fe3O4 + 1/2 O2---> 2Fe2O3
Thus as shown in the balanced equation two moles of Fe2O3 are formed when 0.5 moles of O2 reacted with mixture of FeO and Fe3O4
moles of Fe2O3 = MAss / Molar mass = 4.141 / 159.69 = 0.0259 moles
So moles of O2 needed = 0.5 X 0.0259 = 0.01295
Mass of O2 = moles X molar mass = 0.01295 X 32 = 0.4144 grams
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