Question #68eca
1 Answer
Explanation:
Let's first write the chemical equation for this precipitation reaction:
To find the moles of each substance present initially, we can use the molarity formula:
#"mol"_ ("Pb(NO"_3")"_2) = (0.806"mol"/(cancel("L")))(0.02150cancel("L soln"))#
#= 0.0173# #"mol Pb(NO"_3")"_2#
#"mol"_ ("KCl") = (0.683"mol"/(cancel("L")))(0.01800cancel("L")) = 0.0123# #"mol KCl"#
(I converted the volumes to liters, as is necessary when using this equation.)
To calculate the amount of precipitate that forms, we must first calculate which reactant is limiting. To find this, we simply divide these calculated mole values by the coefficient of the substance in the equation, and whichever number is lower, that reagent is limiting:

#"Pb(NO"_3")"_2# :#(0.0173"mol")/(1 "(coefficient)") = color(red)(0.0173# 
#"KCl"# :#(0.0123"mol")/(2 "(coefficient)") = color(blue)(0.00615#
Therefore,
To find the number of moles of
Thus,
The moles of precipitate (
The amount of moles of precipitate expected is the same amount as lead(II) nitrate used, because their coefficients are both