According to the following reaction,what mass of PbCl₂ can form from 235 mL of 0.110 M KCl solution? Assume that there is excess Pb(NO₃)₂. 2 KCl(aq)+ Pb(NO₃)₂(aq)→ PbCl₂(s)+ 2 KNO₃(aq) A) 7.19 g B) 3.59 g C) 1.80 g D) 5.94 g E) 1.30 g

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Quizplus · Accepted Answer

First, calculate the moles of KCl: $0.235 \, \text{L} \times 0.110 \, \text{mol/L} = 0.02585 \, \text{mol}$. According to the balanced equation, 2 moles of KCl produce 1 mole of PbCl2, so $0.02585 \, \text{mol} \, \text{KCl} \times \frac{1 \, \text{mol} \, \text{PbCl2}}{2 \, \text{mol} \, \text{KCl}} = 0.012925 \, \text{mol} \, \text{PbCl2}$. The molar mass of PbCl2 is approximately 278.1 g/mol, so $0.012925 \, \text{mol} \times 278.1 \, \text{g/mol} = 3.59 \, \text{g}$.

According to the Following Reaction,what Mass of PbCl2 Can Form

According to the Following Reaction,what Mass of PbCl₂ Can Form