Question

calculate the $k_{eq}$ from the given concentrations. $mgcl_2$ and $k_2co_3$ were dissolved and the reaction happened because a solid was formed.
\

$$\begin{tabular}{lcccccc} & $mgcl_{2\\ (aq)}$ & + & $k_2co_{3\\ (aq)}$ & $\ ightleftharpoons$ & $mgco_{3\\ (s)}$ & + $2\\ kcl_{(aq)}$ \\ start: & 1.00 m & & 1.00 m & & 0 & 0 \\ equil: & 0.300 m & & 0.300 m & & n/a & 1.40 m \\ & 0.700 m & & 0.700 m & & & \\ \\end{tabular}$$

first put in the chemical symbols, then put in the numbers. use the order given in the reaction.
the concentration of kcl was squared due to its coefficient.
$k_{eq}^f = \frac{\\_\\_1\\_\\_^2}{\\_\\_4\\_\\_\\ \\_\\_5\\_\\_} = \frac{(\\_\\_2\\_\\_)^2}{(\\_\\_6\\_\\_)(\\_\\_7\\_\\_)} = \\_\\_3\\_\\_

a. $h_3o^{+1}$ b. $oh^{-1}$ c. $k_2co_3$ d. $mgco_3$ e. $mgcl_2$ f. $kcl$ f. $so_3^{-2}$
g. $h_2so_3$ h. $hc_2h_3o_2$ i. $c_2h_3o_2^{-1}$ j. $h_2s$ k. $hs^{-1}$ l. $s^{-2}$
m. $hx$ n. $x^{-1}$ o. 1 p. 2 q. 3 r. 0.5000 s. 0.300 t. 0.333
u. 1.40 v. $1.75 \times 10^{-6}$ w. $1.11 \times 10^{-8}$ x. $5.5 \times 10^{-2}$ y. 0.8000
z. $1.60 \times 10^{-5}$ aa. $1.143 \times 10^{-5}$ bb. 0.4900 cc. $1.104 \times 10^{-5}$
dd. 0.700 ee. $1.26 \times 10^{-5}$ ff. 4.901 gg. 21.8 hh. 4.942 ii. 4.957

Explanation:

Step1: Recall Equilibrium Expression

For the reaction \(\ce{MgCl2(aq) + K2CO3(aq)
ightleftharpoons MgCO3(s) + 2KCl(aq)}\), solids are not included in \(K_{eq}\) (activity of solid is 1, so we ignore it). The equilibrium expression is \(K_{eq}=\frac{[\ce{KCl}]^2}{[\ce{MgCl2}][\ce{K2CO3}]}\).

Step2: Identify Concentrations

From the table, at equilibrium: \([\ce{KCl}] = 1.40\space M\) (option U), \([\ce{MgCl2}] = 0.300\space M\) (option S), \([\ce{K2CO3}] = 0.300\space M\) (option S). Wait, no, wait the equil row: \(\ce{MgCl2}\) equil is \(0.300\space M\), \(\ce{K2CO3}\) equil is \(0.300\space M\), \(\ce{KCl}\) equil is \(1.40\space M\).

Step3: Plug into Expression

Substitute into \(K_{eq}=\frac{[\ce{KCl}]^2}{[\ce{MgCl2}][\ce{K2CO3}]}\). So numerator: \((1.40)^2\), denominator: \((0.300)(0.300)\).

Step4: Calculate Numerator and Denominator

Numerator: \(1.40^2 = 1.96\). Denominator: \(0.300\times0.300 = 0.09\).

Step5: Compute \(K_{eq}\)

\(K_{eq}=\frac{1.96}{0.09}\approx21.78\approx21.8\) (option GG). Wait, wait, no, wait the equil concentrations: Wait the start and equil: Wait the reaction: \(\ce{MgCl2}\) starts at 1.00, equil at 0.300, so change is 0.700. \(\ce{K2CO3}\) starts at 1.00, equil at 0.300, change 0.700. \(\ce{KCl}\) starts at 0, equil at 1.40, which is \(2\times0.700 = 1.40\), correct. So the equilibrium expression is \(K_{eq}=\frac{[\ce{KCl}]^2}{[\ce{MgCl2}][\ce{K2CO3}]}\). So \([\ce{KCl}] = 1.40\) (U), \([\ce{MgCl2}] = 0.300\) (S), \([\ce{K2CO3}] = 0.300\) (S). Wait, no, wait the equil row: \(\ce{MgCl2}\) equil is \(0.300\), \(\ce{K2CO3}\) equil is \(0.300\), \(\ce{KCl}\) equil is \(1.40\). So plugging in: \(\frac{(1.40)^2}{(0.300)(0.300)}=\frac{1.96}{0.09}\approx21.78\approx21.8\) (option GG). Wait, but let's check the options. Wait maybe I misread the equil concentrations. Wait the table: start: \(\ce{MgCl2}\) 1.00 M, \(\ce{K2CO3}\) 1.00 M, \(\ce{MgCO3}\) 0, \(\ce{KCl}\) 0. equil: \(\ce{MgCl2}\) 0.300 M, \(\ce{K2CO3}\) 0.300 M, \(\ce{MgCO3}\) N/A, \(\ce{KCl}\) 1.40 M. So the expression is \(K_{eq}=\frac{[\ce{KCl}]^2}{[\ce{MgCl2}][\ce{K2CO3}]}\). So substituting: \(\frac{(1.40)^2}{(0.300)(0.300)}=\frac{1.96}{0.09}\approx21.78\approx21.8\) (option GG).

Wait, but let's check the blanks:

\([\_1\_]^2\): this is \(\ce{KCl}\) (option F), so \([\ce{KCl}] = 1.40\) (option U), so \((\_2\_)^2\) is (U)²? Wait no, the first blank \([\_1\_]\) is \(\ce{KCl}\) (F), then \((\_2\_)\) is the concentration of \(\ce{KCl}\), which is 1.40 (U). Then denominator: \([\_4\_]\) is \(\ce{MgCl2}\) (E), concentration 0.300 (S); \([\_5\_]\) is \(\ce{K2CO3}\) (C), concentration 0.300 (S). Then \(K_{eq}=\frac{(1.40)^2}{(0.300)(0.300)}=\frac{1.96}{0.09}\approx21.8\) (GG).

Answer:

1. F (KCl)
2. U (1.40)
3. GG (21.8)
4. E (MgCl₂)
5. C (K₂CO₃)
6. S (0.300)
7. S (0.300)

Wait, but the question's blanks: \([\_1\_]^2\) is the chemical symbol for KCl (F), then \((\_2\_)^2\) is the concentration of KCl (U). Then denominator: \([\_4\_]\) is MgCl₂ (E), concentration (\_6\_) is 0.300 (S); \([\_5\_]\) is K₂CO₃ (C), concentration (\_7\_) is 0.300 (S). Then \(K_{eq}=\frac{(1.40)^2}{(0.300)(0.300)} = \frac{1.96}{0.09}\approx21.8\) (GG). So the final \(K_{eq}\) is 21.8 (GG).