How many grams of pbbr2 will precipitate when excess crbr; solution is added to 60.0 ml of 0.551 m pb(no3)2 solution?
3pb(no3)2(aq) + 2crbr3(aq) –> 3pbbrz(s) + 2cr(no3)(aq)

The mass of PbBr2 precipitate produced is 12.1 g.

The equation of the reaction is;

3Pb(NO3)2(aq) + 2CrBr3(aq) –>3PbBr2(s) + 2Cr(NO3)3(aq)

Number of moles of Pb(NO3)2 =  60.0/1000 L ×  0.551 M = 0.0331 moles

From the information in the question, we already know that  Pb(NO3)2 is the limiting reactant and  CrBr3 is the reactant in excess.

From the equation of the reaction;

3 moles of Pb(NO3) yields 3 moles of PbBr2

0.0331 moles of Pb(NO3) yields 0.0331 moles ×  3 moles/3 moles

=  0.0331 moles of PbBr2

Mass of PbBr2 = number of moles × molar mass

Mass of PbBr2 =  0.0331 moles × 367 g/mol= 12.1 g

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5.48 g

Explanation:

M:                                                     367.01

    3Pb(NO3)2(aq) + 2CrBr3(aq) → 3PbBr2(s) + 2Cr(NO3)3(aq)

1. Moles of Pb(NO3)2

n = 30.0 mL × (0.551 mmol/1 mL) = 16.53 mmol Pb(NO3)2

2. Moles of PbBr2

n = 16.53 mmol Pb(NO3)2 × (3 mmol PbBr/3 mol Pb(NO3)2)

= 16.53 mmol PbBr2

(3) Mass of PbBr2 formed

m = 16.53 mmol × (367.01 mg/1 mol) = 6067 mg

(4) Mass of PbBr2 in solution

PbBr2 is slightly soluble, so a significant amount will remain in solution.

Its solubility is 973 mg/100 mL.

Mass of dissolved PbBr2 = 60.0 mL × 973 mg/100 mL = 584 mg

(5) Mass of precipitate

Mass of precipitate = mass of PbBr2 formed - mass of PbBr2 in solution

                            m = 6067 mg - 584 mg = 5480 mg = 5.48 g

The mass of precipitate is 5.48 g.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year