the balanced equation is Pb(NO3)2 (aq) + 2KCl(aq) PbCl2(s) + 2KNO3(aq) convert g

Question

the balanced equation is

Pb(NO3)2 (aq) + 2KCl(aq) PbCl2(s) + 2KNO3(aq)

convert grams to moles by using molar mass

(7.51 Pb(NO3)2 ) / ( 331.2098 g Pb(NO3)2 / mol) = 0.0227 mol Pb(NO3)2

( 7.08 g KCl ) / ( 74.5513 g KCl / mol) = 0.0949 mol KCl

0.0227 mol Pb(NO3)2 would react completely with 0.0227 * (2/1) = 0.0454 moles of KCl but more KCl present than that so KCl is excess and  Pb(NO3)2 is the limiting reactant

( 0.0227 mol Pb(NO3)2 ) * ( 1 mol PbCl2 / 1 mol Pb(NO3)2 ) * ( 278.106 g PbCl2 / mol) *( 0.898) = 5.66 g PbCl2 recovered

supoosing the less than 100% recovery was not caused by an incomplete reaction

( 0.0949 mol KCl initially ) - ( 0.0454 moles of KCl reacted) * ( 74.5513 g KCl / mol) = 3.69 g KCl left over

Explanation / Answer

Sapling Learning Map An aqueous solution containing 7.51 g of leadl) nitrate is added to an aqueous solution containing 7.08 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(II) nitrate O potassium chloride The percent yield for the reaction is 89 896, how many grams of precipitate were recovered? Number How many grams of the excess reactant remain? Number O Previous Check Answer 0 Next Ext Hint about us careers privacy policy terms of use contact us help

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