Question

If 34.56 mL of 0.500 M copper(II) nitrate reacts with 47.89 mL of 0.600 M sodium hydroxide, what is the limiting reactant?

a) Copper(II) nitrate
b) Sodium hydroxide
c) Both reactants are in stoichiometric proportions
d) Cannot be determined from the given information

Answer 1

The limiting reactant between copper(II) nitrate and sodium hydroxide is sodium hydroxide (NaOH) since it is available in a lesser amount than needed to fully react with the supplied copper(II) nitrate.a) Copper(II) nitrate

Step-by-step explanation:

To determine the limiting reactant of the reaction between copper(II) nitrate and sodium hydroxide, we must first calculate the moles of each reactant involved. We use the following formulas:

• Moles of copper(II) nitrate = Volume (L) × Molarity (M)
• Moles of sodium hydroxide = Volume (L) × Molarity (M)

For copper(II) nitrate:

0.03456 L × 0.500 M = 0.01728 mol Cu(NO3)2

For sodium hydroxide:

0.04789 L × 0.600 M = 0.028734 mol NaOH

The balanced equation for the reaction is:

Cu(NO3)2 + 2 NaOH → Cu(OH)2 + 2 NaNO3

From this equation, we see that 1 mol of Cu(NO3)2 reacts with 2 mol of NaOH. So, for the 0.01728 mol of Cu(NO3)2, we need 2 × 0.01728 mol = 0.03456 mol of NaOH. Since we have 0.028734 mol NaOH, it's less than the required 0.03456 mol. Therefore, the limiting reactant is sodium hydroxide (NaOH).