Answer: 16.32 g of
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of
require = 1 mole of
Thus 0.34 moles of
will require=
of
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of
left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of
as excess reagent are left.
The number of carbon atoms in an alcohol affects its solubility in water, as shown in Table 13.3. As the length of the carbon chain increases, the polar OH group becomes an ever smaller part of the molecule, and the molecule becomes more like a hydrocarbon. The solubility of the alcohol decreases correspondingly.
132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!
Double replacement because H and K are both switching