Question

If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, what is the limiting reactant?

Answer 1

Answer:

Pb(NO3)2- Lead Nitrate

NaI - Sodium Iodide

P

b

(

N

O

3

)

2

+

2

N

a

I

−

→

P

b

I

2

+

2

N

a

N

O

3

Moles of Lead(II) Nitrate =

m

a

s

s

/

m

o

l

a

r

m

a

s

s

= 25.0 grams//(207.20 xx 1 + 14.01 xx 2 + 16 xx 6

= 25.0 grams//331.22

= 0.075 moles

Moles of Sodium Iodide

= 15//(22.99+126.9)

= 15//149.89

= 0.100 Moles

Limiting Reagent in this case is Lead(II) Nitrate

Ratio of Moles of Lead Nitrate to Sodium Nitrate

1:2

Ratio of "moles" of lead nitrate to Sodium Nitrate

0.075 : x

1

2

=

0.075

x

x

=

2

×

0.075

x

=

0.150

Thus that means 0.150 moles of Sodium Nitrate

To calculate mass of Sodium Nitrate

Since we know that Moles =

M

a

s

s

/

M

o

l

a

r

M

a

s

s

Thus Mass =

M

o

l

e

s

×

M

o

l

a

r

M

a

s

s

Molar Mass of

N

a

N

O

3

Na - 22.99

N - 14.01

O - 3(16) = 48

85g/mol

Thus, Mass =

0.150

×

85

Mass =

12.75

Thus, 12.75g of Sodium Nitrate can be formed