If 0.7825 g of solid Al and 450. mL of 0.2069 M aqueous Cl- are reacted stoichiometrically according to the equation, what mass (g) of solid Al remained?
3 CuCl2(aq) + 2 Al(s) �¨ 3 Cu + 2 AlCl3(aq)
To find the mass of solid Al that remains after the reaction, you need to determine the limiting reactant first.
Step 1: Calculate the number of moles of Al
Given: Mass of Al = 0.7825 g
To calculate the number of moles, use the molar mass of Al, which is 26.98 g/mol.
Number of moles of Al = (0.7825 g) / (26.98 g/mol) = 0.02899 mol
Step 2: Calculate the number of moles of CuCl2
Given: Volume of CuCl2 solution = 450 mL = 450/1000 L = 0.45 L
Given: Concentration of CuCl2 solution = 0.2069 M
To calculate the number of moles, use the formula: Moles = Volume x Concentration.
Number of moles of CuCl2 = (0.45 L) x (0.2069 mol/L) = 0.09261 mol
Step 3: Determine the stoichiometry of the balanced equation
The balanced equation states that 3 moles of CuCl2 react with 2 moles of Al. So, the ratio is 3:2.
Step 4: Compare the moles of the two reactants to find the limiting reactant
By comparing the moles of CuCl2 (0.09261 mol) and Al (0.02899 mol), we can see that Al is the limiting reactant. This means that Al is completely consumed in the reaction, and some CuCl2 is left over.
Step 5: Calculate the mass of Al remaining
Since all the Al is consumed, there is no Al remaining after the reaction. Therefore, the mass of solid Al remaining is 0 g.
Answer: The mass of solid Al remaining is 0 g.