Question

Student A- 5.01g NaCl(s) reacts with 200mL of 0.35M solution of AgNO3 (aq)
)with percentage yield of 76.3%.
Student B- 6.58g AgNO3(aq) reacts with 40mL of 1.45M solution of BaCl2(aq) with percentage yield of 68.5%.
What mass of silver chloride solid is produced by each student and what happens when the concentration of the solutions used was doubled?

Answer 1

Answer on Question #72541, Chemistry / General Chemistry

Student A- 5.01g NaCl(s) reacts with 200mL of 0.35M solution of $\mathrm{AgNO_3}$ (aq) with percentage yield of 76.3%. Student B- 6.58g $\mathrm{AgNO_3}$(aq) reacts with 40mL of 1.45M solution of $\mathrm{BaCl_2}$(aq) with percentage yield of 68.5%. What mass of silver chloride solid is produced by each student and what happens when the concentration of the solutions used was doubled?

Solution

Student A:

$$\mathrm{NaCl} + \mathrm{AgNO_3} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}$$

As it's seen from the equation 1 mole of NaCl reacts with 1 mole of $\mathrm{AgNO_3}$ producing 1 mole of AgCl.

Find the amounts of the reactants:

$$\begin{array}{l} v(\mathrm{NaCl}) = \frac{m}{M} = \frac{5.01}{58.5} = 0.086 \text{ (mole)} \\ v(\mathrm{AgNO_3}) = C_m \times V = 0.35 \times 0.2 = 0.07 \text{ (mole)} \end{array}$$

As NaCl is in excess then there is produced 0.07 mole of AgCl. Find it's mass:

$$\begin{array}{l} m(\mathrm{AgCl}) = M \times v = 143 \times 0.07 = 10.01 \text{ (g)} - \text{theoretical yield}; \\ m_{\text{pr}}(\mathrm{AgCl}) = m_{\text{theor}} \times w = 10.01 \times 0.763 = 7.64 \text{ (g)} - \text{mass of solid AgCl is produced by student A}. \end{array}$$

If the concentration of the solution used was doubled, then $\mathrm{AgNO_3}$ would become in excess, so the theoretical yield of AgCl would be 0.086 mole instead of 0.07 mole.

Student B:

$$\mathrm{BaCl_2} + 2\mathrm{AgNO_3} \rightarrow 2\mathrm{AgCl} + \mathrm{Ba(NO_3)_2}$$

As it's seen from the equation 1 mole of $\mathrm{BaCl_2}$ reacts with 2 moles of $\mathrm{AgNO_3}$ producing 2 moles of AgCl.

Find the amounts of the reactants:

$$\begin{array}{l} v(\mathrm{AgNO_3}) = \frac{m}{M} = \frac{6.58}{170} = 0.039 \text{ (mole)} \\ v(\mathrm{BaCl_2}) = C_m \times V = 1.45 \times 0.04 = 0.058 \text{ (mole)} \end{array}$$

0.039 mole of $\mathrm{AgNO_3}$ requires 0.0195 mole of $\mathrm{BaCl_2}$. So $\mathrm{BaCl_2}$ is in excess and there is produced 0.039 mole of AgCl. Find it's mass:

$$\begin{array}{l} m(\mathrm{AgCl}) = M \times v = 143 \times 0.039 = 5.58 \text{ (g)} - \text{theoretical yield}; \\ m_{\text{pr}}(\mathrm{AgCl}) = m_{\text{theor}} \times w = 5.58 \times 0.685 = 3.82 \text{ (g)} - \text{mass of solid AgCl is produced by student B}. \end{array}$$

The doubling of the concentration of $\mathrm{BaCl}_2$ solution leads to increasing of $\mathrm{BaCl}_2$ amount. As it is already in excess, so nothing happens. The theoretical yield of AgCl is still 0.039 mole.

**Answer**

Student A: $m(\mathrm{AgCl}) = 7.64\ \mathrm{g}$

Student B: $m(\mathrm{AgCl}) = 3.82\ \mathrm{g}$

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