Question

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

Answer 1

To determine the grams of lead (II) chloride that can be collected, calculate the moles of sodium chloride and lead (II) chloride using the balanced chemical equation and molar masses.

Step-by-step explanation:

To determine the grams of lead (II) chloride that can be collected, we need to calculate the moles of sodium chloride and lead (II) chloride. First, calculate the moles of sodium chloride using its molar mass. Then, use the balanced chemical equation to determine the mole ratio between sodium chloride and lead (II) chloride. Finally, convert the moles of lead (II) chloride to grams using its molar mass.

The balanced chemical equation for the reaction is:

Pb(NO3)2 + 2 NaCl -> PbCl2 + 2 NaNO3

Answer 2

119.9 grams of Lead chloride will be collected.

Step-by-step 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.