When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

And the resulting percent yield:

Regards!

the dependant variable is spilled water

the independent is the heat

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.