Anhydrous CuSO4 can be used to dry liquids in which it is insoluble. The CuSO4 is converted to CuSO4 x 5H20, which can be filtered off from the liquid. What is the minimum mass of anhydrous CuSO4 needed to remove 12.6 g H20 from a tankful of gasoline?
RMM of CuSO4=(63.55+32+4*16)=159.55
RMM of 5H2O = 5*(1*2+16)=90
Use proportions,
Amount of CuSO4 required
=12.6g *(RMM CuSO4)/(RMM 5H2O)
=?
1*2+16=18
5*18=90
To determine the minimum mass of anhydrous CuSO4 needed to remove 12.6 g of water from the tankful of gasoline, we need to consider the molar mass and stoichiometry of CuSO4⋅5H2O.
Step 1: Determine the molar mass of CuSO4⋅5H2O
The molar mass of CuSO4 is approximately 159.61 g/mol.
The molar mass of H2O is approximately 18.02 g/mol.
Since there are five water molecules (5H2O) in CuSO4⋅5H2O, the molar mass of CuSO4⋅5H2O is:
Molar mass of CuSO4⋅5H2O = (159.61 g/mol) + (5 * 18.02 g/mol)
Molar mass of CuSO4⋅5H2O = 249.61 g/mol
Step 2: Calculate the moles of water present in the tankful of gasoline
Moles of water = given mass / molar mass
Moles of water = 12.6 g / 18.02 g/mol
Moles of water = 0.699 mol (approximately)
Step 3: Use stoichiometry to relate moles of water to moles of anhydrous CuSO4
From the balanced chemical equation:
CuSO4⋅5H2O → CuSO4 + 5H2O
1 mole of CuSO4⋅5H2O will produce 5 moles of water.
Therefore, 0.699 moles of water will require (0.699 mol / 5) = 0.14 moles of anhydrous CuSO4.
Step 4: Calculate the mass of anhydrous CuSO4
Mass of anhydrous CuSO4 = moles of anhydrous CuSO4 * molar mass of anhydrous CuSO4
Mass of anhydrous CuSO4 = 0.14 mol * 159.61 g/mol
Mass of anhydrous CuSO4 = 22.34 g
Therefore, the minimum mass of anhydrous CuSO4 needed to remove 12.6 g of water from the tankful of gasoline is approximately 22.34 grams.
To find the minimum mass of anhydrous CuSO4 needed to remove 12.6 g H2O from the gasoline, we need to calculate the amount of CuSO4 x 5H2O needed to absorb that amount of water.
The molar mass of H2O is 18 g/mol.
We know that copper sulfate pentahydrate (CuSO4 x 5H2O) has a molar mass of 249.5 g/mol.
To calculate the minimum mass of anhydrous CuSO4, we can use the following formula:
Mass of anhydrous CuSO4 = (mass of water to be removed) * (molar mass of CuSO4 x 5H2O) / (molar mass of water)
Plugging in the values:
Mass of anhydrous CuSO4 = 12.6 g * (249.5 g/mol) / (18 g/mol)
Simplifying the equation:
Mass of anhydrous CuSO4 = 174.75 g
Therefore, the minimum mass of anhydrous CuSO4 needed to remove 12.6 g H2O from the gasoline is 174.75 grams.