Answer:
88.98 %.
Explanation:
- From the balanced equation:
<em>2HCl + Pb(NO₃)₂ → 2HNO₃ + PbCl₂</em>
- It is clear that 1.0 mole of Pb(NO₃)₂ reacts with 2.0 moles of HCl to produce 1.0 mole of PbCl₂ and 2.0 moles of HNO₃.
- <em>The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100.</em>
- The actual yield of lead(II) chloride (PbCl₂) = 650 g.
- Now, we need to calculate the calculated yield of lead(II) chloride (PbCl₂).
- We need to calculate the no. of moles (n) of lead(II) nitrate (Pb(NO₃)₂) (870 grams) using the relation: <em>n = mass / molar mass.</em>
- n of lead(II) nitrate (Pb(NO₃)₂) = mass / molar mass = (870 g) / (331.2 g/mol) = 2.63 mol.
- Since HCl is in excess, the limiting reactant is lead(II) nitrate (Pb(NO₃)₂).
<u><em>Using cross multiplication:</em></u>
1.0 mole of Pb(NO₃)₂ produces → 1.0 mole of PbCl₂, from the stichiometry.
∴ 2.63 mole of Pb(NO₃)₂ produces → 2.63 mole of PbCl₂.
- The mass of PbCl₂ produced (the calculated yield) = n x molar mass = (2.63 mol) (278.1 g/mol) = 730.52 g.
∴ The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100 = [(650 g) / (730.52)] x 100 = 88.98 %.
Answer:
Independent variable: different alcohols
Dependent variable: red dye
