How many grams of lead (2) chloride can be collected from 34.5 g sodium chloride and an excess amount of lead (2) nitrate?

119.9 grams of Lead chloride will be collected.

Explanation:

PbNO3 + 2NaCl PbCl2+2NaNO3

From the equation we know that 2 moles of NaCl is required to convert 1 mole of PbNO3 to PbCl2.

Lead nitrate here, is a limiting agent:

No of moles of NaCl used is calculated as the mass is given, atomic mass of NaCl is 40

No. of moles= wt/atomic weight

                       =  34.5/40

                        = 0.8625 moles of Nacl will be required.

So from stoichiometry,

1/2=  x/0.862      

0.862=2x

x=0.862/2

 = 0.431 moles

So x is moles of PbCl2 precipitated

From the formula

n=wt/at wt

wt= n*at wt

    = 0.431*278.1  ( 278.1 is the atomic weight of lead chloride)

     = 119.9 grams of PbCl2

lead nitrate required will be 1 mole of Lead nitrate amounts to 331.2 gms.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )