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Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
*
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
*
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be:
- 5.76 - 3.86 = 1.9 g KCl