Anhydrous CuSO4 can be used to dry liquids in which it is insoluble. The CuSO4 is converted to CuSO4.5H2O, which can be filtered off from the liquid.
What is the minimum mass of anhydrous CuSO4 needed to remove 12.4g H2O from a tankful of gasoline?
how many moles of H2O are in 12.4g?
moles H2O = 12.4/18 grams
so to get CuSO4.5H2O, you need (12.4/18)*1/5 moles of CuSO4
figure how many grams that is.
To find the minimum mass of anhydrous CuSO4 needed to remove 12.4g H2O from a tankful of gasoline, we need to use stoichiometry and consider the molar ratio between CuSO4.5H2O and H2O.
The molar mass of H2O is 18.015 g/mol, and the molar mass of CuSO4.5H2O is 249.68 g/mol.
The stoichiometry indicates that there is 1 mol of H2O for every 1 mol of CuSO4.5H2O. Therefore, we can set up a ratio as follows:
(1 mol CuSO4.5H2O / 1 mol H2O) = (249.68 g CuSO4.5H2O / 18.015 g H2O)
Now, let's calculate the moles of H2O in 12.4g:
moles H2O = mass H2O / molar mass H2O
moles H2O = 12.4 g / 18.015 g/mol
Next, we can use the ratio to find the moles of CuSO4.5H2O required:
moles CuSO4.5H2O = moles H2O x (1 mol CuSO4.5H2O / 1 mol H2O)
Finally, to find the mass of anhydrous CuSO4, which has the same number of moles as CuSO4.5H2O:
mass CuSO4 = moles CuSO4.5H2O x molar mass CuSO4.5H2O
Let's calculate it:
moles CuSO4 = (12.4 g / 18.015 g/mol) x (1 mol CuSO4.5H2O / 1 mol H2O) = 0.686 mol CuSO4.5H2O
mass CuSO4 = 0.686 mol CuSO4.5H2O x 249.68 g/mol = 171.2 g
Therefore, the minimum mass of anhydrous CuSO4 needed to remove 12.4g H2O from a tankful of gasoline is approximately 171.2 grams.