Here’s what I got.

Since you didn’t provide the concentration of the lead(II) nitrate solution, we’re going to have to improvise a bit here.

You’re dealing with a in which aqueous potassium chromate, ##”K”_2″CrO”_4##, and aqueous lead(II) nitrate, ##”Pb”(“NO”_3)_2##, will react to form the insoluble lead(II) chromate, ##”PbCrO”_4##, and aqueous potassium nitrate, ##”KNO”_3##.

The balanced chemical equation that describes this reaction looks like this

##”K”_ 2″CrO”_ (4(aq)) + “Pb”(“NO”_ 3)_ (2(aq)) -> “PbCrO”_ (4(s)) darr + 2″KNO”_(3(aq))##

Notice that the two reactants react in a ##1:1## . This tells you that the reaction will *always consume equal numbers of moles of potassium chromate and of lead(II) nitrate.

Moreover, lead(II) chromate is produced in the same ##1:1## mole ratio with both reactants, which means that the number of moles of lead(II) chromate produced by the reaction will match the number of moles of the reactants that take part in the reaction.

Use the and volume of the potassium chromate solution to find how many moles of this reactant you have present

##color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_”solute”/V_”solution” implies n_”solute” = c * V_”solution”)color(white)(a/a)|)))##

In your case, you have

##n_(K_2CrO_4) = “0.40 mol” color(red)(cancel(color(black)(“L”^(-1)))) * overbrace(15.0 * 10^(-3)color(red)(cancel(color(black)(“L”))))^(color(blue)(“volume in liters”))##

##n_(K_2CrO_4) = “0.00600 moles K”_2″CrO”_4##

Now, you have three possible scenarios to account for here

##color(white)(a)##

In this case, lead(II) nitrate will be the because it will be completely consumed by the reaction before all the moles of potassium chromate will get the chance to react.

##n_(Pb(NO_3)_2) < “0.00600 moles”##

Since you have equal volumes of both , you can say that if the concentration of the lead(II) nitrate is lower than that of potassium chromate, the former will act as a .

##color(green)(|bar(ul(color(white)(a/a)color(black)(c_(Pb(NO_3)_2) color(red)(## lead(II) nitrate will be the limiting reagent

##color(white)(a)##

In this case, both of the reactants will be completely consumed by the reaction, i.e. you won’t be dealing with a limiting reagent.

Since the lead(II) chromate is produced in a ##1:1## mole ratio with the two reactants, it follows that the reaction will produce ##0.00600## moles of ##”K”_2″CrO”_4##.

Use the compound’s molar mass to find how many Grams would contain this many moles

##0.00600 color(red)(cancel(color(black)(“moles K”_2″CrO”_4))) * “184.19 g”/(1color(red)(cancel(color(black)(“mole K”_2″CrO”_4)))) = color(green)(|bar(ul(color(white)(a/a)”1.1 g “color(white)(a/a)|)))##

##color(white)(a)##

This time, the potassium chromate will be completely consumed by the reaction before all the moles of lead(II) nitrate get the chance to react.

Potassium chromate will act as a limiting reagent. The reaction will once again produce ##0.00600## moles of lead(II) chromate. So, for

##color(green)(|bar(ul(color(white)(a/a)color(black)(c_(Pb(NO_3)_2) color(red)(>) “0.40 mol L”^(-1))color(white)(a/a)|))) ->## potassium chromate will be the limiting reagent

You will have

##0.00600 color(red)(cancel(color(black)(“moles K”_2″CrO”_4))) * “184.19 g”/(1color(red)(cancel(color(black)(“mole K”_2″CrO”_4)))) = color(green)(|bar(ul(color(white)(a/a)”1.1 g “color(white)(a/a)|)))##

The answers are rounded to two .

It’s worth noting that lead(II) chromate is a yellow insoluble solid that precipitates out of solution as the reaction proceeds.

Our website has a team of professional writers who can help you write any of your homework. They will write your papers from scratch. We also have a team of editors just to make sure all papers are of HIGH QUALITY & PLAGIARISM FREE. To make an Order you only need to click Ask A Question and we will direct you to our Order Page at WriteDemy. Then fill Our Order Form with all your assignment instructions. Select your deadline and pay for your paper. You will get it few hours before your set deadline.