Final answer:
Upon mixing 1250 mL of 0.0500 M CuSO4 with 2000 mL of 0.0250 M NaOH, 2.439 grams of Cu(OH)2 precipitate is produced, with NaOH being the limiting reactant.
Step-by-step explanation:
When 1250 mL of 0.0500 M CuSO4 and 2000 mL of 0.0250 M NaOH are mixed, the reaction between them will produce copper hydroxide (Cu(OH)2), which is a precipitate. The balanced chemical equation for the reaction is:
CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
To calculate the mass of the precipitate, first determine the limiting reactant. The number of moles of each reactant is:
- CuSO4: 1.25 L x 0.0500 mol/L = 0.0625 mol
- NaOH: 2.00 L x 0.0250 mol/L = 0.0500 mol
Based on the stoichiometry of the reaction, 1 mole of CuSO4 reacts with 2 moles of NaOH. Hence, NaOH is the limiting reactant.
The mass of Cu(OH)2 produced can be calculated from the moles of NaOH:
- Number of moles of Cu(OH)2: 0.0250 mol (because it reacts in a 1:2 ratio with NaOH)
- Molar mass of Cu(OH)2: 97.561 g/mol
- Mass of Cu(OH)2: 0.0250 mol x 97.561 g/mol = 2.439 g
Therefore, the mass of the precipitate formed is 2.439 grams of Cu(OH)2.