5. 34.1 grams of lead(II) nitrate are reacted with 4.55 grams of sodium chloride, producing lead(II) chloride and sodium artant are consumed in trie nitrate. 1 Po(NO3+Nacl=Pbclz tANa NOz 1.19 PONO3 a Pla-o.103 or -20 imol Pbl WANO3 1.5g Naci (- | mol Nach 8.445 Naci T ol Po(N la =0.0779/0r.156 reactant initing The sodium chloride will be limiting reactant a. Which reactant will be limiting? b. Determi the mass of each product. Imol Pb(NO Pbda 1 Pb(NO 278.19 Pb(), Imol Pbcl2 33 Ph- 207.20 Pbl2 %3D Cl-2(35.45) ちのge 278.1 2 NANO3 Pa(ND I Po(NOS) Na No Imol Da NOz Na No3 34.1 Ph(NO3).(mol POCNDSJA Na-22.99 N-14.01 =85 35g sh = 0

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34.1 grams of lead(II) nitrate are reacted with 4.55 grams of sodium chloride, producing lead(II) chloride and sodium nitrate. Which reactant will be limiting? Determine the mass of each product.

5. 34.1 grams of lead(II) nitrate are reacted with 4.55 grams of sodium chloride, producing lead(II) chloride and sodium
actant are consumed in
nitrate.
1 Po(NO3+Nacl=PbCl2 tNa NOz
c.
Imol Pbli
WaNO3
I hol Pb(Ne
-0.103 or.206
9.59 Naci
Naci / molNach
58.
.44g Nac
NANO3
=0.0779/6r.156
I mot Nacl
limiting reactant
imiting
a. Which reactant will be limiting?
The sodium chloride will be limiting reactant
b. Determine the mass of each product.
1.
Pbl2
278.19 PbC),
Imol Pbcl2
Imol Pb(NOS)
34.
.19 Po (NO3 331.21q PoCMO
)
1 Pb(NO)
Pb- 207.20
1-2235.45)
Pbcl2
|
28.6
278.1
2 NANO3
9 Na No
I molWaNOz
I mb) Pb(ND3)a
Pb (NO3)g(-
34.1g
331.
3D185
Na No3
Na-22.99
N-14.01
sh D0
85
Transcribed Image Text:5. 34.1 grams of lead(II) nitrate are reacted with 4.55 grams of sodium chloride, producing lead(II) chloride and sodium actant are consumed in nitrate. 1 Po(NO3+Nacl=PbCl2 tNa NOz c. Imol Pbli WaNO3 I hol Pb(Ne -0.103 or.206 9.59 Naci Naci / molNach 58. .44g Nac NANO3 =0.0779/6r.156 I mot Nacl limiting reactant imiting a. Which reactant will be limiting? The sodium chloride will be limiting reactant b. Determine the mass of each product. 1. Pbl2 278.19 PbC), Imol Pbcl2 Imol Pb(NOS) 34. .19 Po (NO3 331.21q PoCMO ) 1 Pb(NO) Pb- 207.20 1-2235.45) Pbcl2 | 28.6 278.1 2 NANO3 9 Na No I molWaNOz I mb) Pb(ND3)a Pb (NO3)g(- 34.1g 331. 3D185 Na No3 Na-22.99 N-14.01 sh D0 85
Expert Solution
Step 1

Given,

Mass of lead (II) nitrate [Pb(NO3)2] reacts = 34.1 g

Mass of sodium chloride (NaCl) reacts = 4.55 g

Limiting reactant = ?

Mass of lead (II) chloride (PbCl2) produced = ?

Mass of sodium nitrate (NaNO3) produced = ?

Note:

Molar mass of lead (II) nitrate [Pb(NO3)2] = (207.2 + 2×14 + 6×16) g/mol = 331.2 g/mol

Molar mass of sodium chloride (NaCl) = (23 + 35.45) g/mol = 58.45 g/mol

Molar mass of lead (II) chloride (PbCl2)= (207.2 + 2×35.45) g/mol = 278.1 g/mol

 Molar mass of sodium nitrate (NaNO3) = (23 + 14 + 3×16) g/mol = 85 g/mol

 

 

 

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