Question

consider the precipitation reaction: bacl₂ + 2 agno₃ → 2 agcl + ba(no₃)₂. how many grams of agcl are generated when 85 g of bacl₂ reacts? round your answer to the correct number of significant figures. bacl₂: 208.23 g/mol agcl: 143.32 g/mol 210 g of agcl 120 g of agcl 58 g of agcl 250 g of agcl 140 g of agcl

Explanation:

Step1: Calculate moles of \( BaCl_2 \)

Moles of \( BaCl_2 = \frac{\text{mass of } BaCl_2}{\text{molar mass of } BaCl_2} = \frac{85\ g}{208.23\ g/mol} \approx 0.4082\ mol \)

Step2: Determine moles of \( AgCl \) from stoichiometry

From the reaction \( BaCl_2 + 2AgNO_3
ightarrow 2AgCl + Ba(NO_3)_2 \), 1 mole of \( BaCl_2 \) produces 2 moles of \( AgCl \). So moles of \( AgCl = 2 \times \) moles of \( BaCl_2 = 2 \times 0.4082\ mol = 0.8164\ mol \)

Step3: Calculate mass of \( AgCl \)

Mass of \( AgCl = \text{moles of } AgCl \times \text{molar mass of } AgCl = 0.8164\ mol \times 143.32\ g/mol \approx 117\ g \). Wait, but let's recalculate more accurately: \( 0.4082\times2 = 0.8164 \); \( 0.8164\times143.32 = 0.8164\times143.32 \approx 117.0 \)? Wait, maybe I made a mistake. Wait, 85 divided by 208.23: 85/208.23 ≈ 0.4081. Then times 2: 0.8162. Then times 143.32: 0.8162143.32 ≈ 0.8162143 = 0.8162140 + 0.81623 = 114.268 + 2.4486 = 116.7166; 0.8162*0.32=0.261184; total ≈ 116.7166+0.261184≈116.9778≈117 g. But the options have 120 g? Wait, maybe the 85 g is considered with two significant figures? Wait, 85 has two significant figures. So let's redo with 85 (two sig figs):

Moles of \( BaCl_2 = 85 / 208.23 ≈ 0.408 \) (but 85 is two sig figs, so maybe 0.41? No, 85 is exact? Wait, the problem says "85 g of \( BaCl_2 \)". So 85 has two significant figures. Then moles: 85 / 208.23 ≈ 0.408, times 2: 0.816, times 143.32: ≈ 117. But the options have 120 g (which is two sig figs, 1.2×10²). Wait, maybe my calculation is wrong. Wait, 208.232=416.46; 143.322=286.64. So the ratio of masses: (mass of AgCl) = (85 g BaCl2) (286.64 g AgCl / 208.23 g BaCl2) = 85 286.64 / 208.23. Let's calculate that: 286.64 / 208.23 ≈ 1.3765. Then 85 1.3765 ≈ 116.99 ≈ 117 g. But the options have 120 g (which is 1.2×10², two sig figs). Wait, maybe the problem expects using approximate values. Wait, 208.23 ≈ 208, 143.32 ≈ 143. Then 85/208 = 0.4086, times 2=0.8172, times 143= 0.8172143= 116.86 ≈ 117. But the options: 210, 120, 58, 250, 140. Wait, 120 is the closest? Wait, maybe I messed up the stoichiometry. Wait, the reaction is 1 mole BaCl2 produces 2 moles AgCl. So molar ratio 1:2. So mass ratio: (2143.32)/208.23 = 286.64/208.23 ≈ 1.376. So 851.376 ≈ 116.96 ≈ 117, which is close to 120 (if we consider significant figures, 85 has two, so 120 (two sig figs)). So the answer should be 120 g of AgCl.

Answer:

120 g of AgCl (the option: 120 g of AgCl)