Question

If you react 2.343 g of sodium hydroxide with excess hydrochloric acid and isolate 2.954 g of sodium chloride, what is your percent yield? The percent yield is what?

Answer 1

Answer 2

Let's solve this problem step-by-step:

1. Write the Balanced Chemical Equation

The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is a neutralization reaction:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

The equation is already balanced.

2. Calculate the Molar Masses

• NaOH: 22.99 (Na) + 16.00 (O) + 1.01 (H) = 40.00 g/mol

• NaCl: 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol

3. Calculate the Theoretical Yield of NaCl

• We are given 2.343 g of NaOH.

• Convert grams of NaOH to moles of NaOH:

  (2.343 g NaOH) / (40.00 g/mol NaOH) = 0.058575 mol NaOH

• From the balanced equation, the mole ratio of NaOH to NaCl is 1:1. Therefore, 0.058575 mol of NaOH will produce 0.058575 mol of NaCl.

• Convert moles of NaCl to grams of NaCl:

  (0.058575 mol NaCl) * (58.44 g/mol NaCl) = 3.423 g NaCl

• So, the theoretical yield of NaCl is 3.423 g.

4. Calculate the Percent Yield

• We are given the actual yield of NaCl as 2.954 g.

• Percent Yield = (Actual Yield / Theoretical Yield) * 100%

• Percent Yield = (2.954 g / 3.423 g) * 100%

• Percent Yield ≈ 86.29%

Answer:

The percent yield is approximately 86.29%.

Answer 3

The percent yield is 86.29 %