Question

An aqueous solution containing 7.30 g of lead(II) nitrate is added to an aqueous solution containing 6.87 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.

An aqueous solution containing 7.30 g of lead(lI) nitrate is added to an aqueous solution containing 6.87 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? O lead(lI) nitrate O potassium chloride The percent yield for the reaction is 89.7%, how many grams of precipitate were recovered? Number How many grams of the excess reactant remain? Number

Answer 1

Pb (NO3)_2 + 2k cl rightarrow Pb cl_2 + 2 K NO_3 According to stoichiometric equation one mole of Pb (NO_3)_2 reacts with 2 moles of KCL to form one mole of Pb cl_2 and 2 moles of KNO_3 have given data no of moles of = 4.14 times of Pb (NO_3)_2 Limiting = Pb (NO_3)_2 Theoretical yield = 0.022 moles of Pbcl_2 formed = 0.022 times 278 .1 /mole = 6.118 g Theoretical yield = 6.118 gr % yield = 89.7% calculated yield =% yield times 6.118 g = 89.7 times 6.118/100 = 5.48 gr calculated yield = 5.48 gr * excess constant 0.0912 moles - 0.022(2) = 0.0912 - 0.044 = 0.0472 moles excess grams of reactant (kcl) = 0.0472 moles times 74.5 gr/moles excess moles of reactant = 3.516 gr