How many moles of lead chloride(II) will be formed from a reaction between 6.5 gram of pbo and 3.2 gram of HCL?
Answers
the equation for the reaction,
Pbo + 2HCL ----------> Pbcl2 + H2O
molecular mass (Pbo) = 207 + 16 = 223 gram
molecular mass(HCl) = 2 * 36.5 = 73 gram
molecular mass(Pbcl2) = 207 + 71 = 278 gram
molecular mass (H2O) = 2 + 16 = 18 gram
From this equation we find the 223 gram of pbo reacts with 73gram of HCL to form 278 gram of pbcl2.
if we carry out the reaction between 3.2 g HCl and 6.5 g PbO.
Amount of PbO that reacts with 3.2 g HCl
= 223 * 3.2 /73
= 9.77g
Since amount of Pbo present is only 6.5 g Pbo is the limiting reagent.
Amount of PbCl2 formed by 6.5 g of PbO
= 278 * 6.5 / 223
No of moles of PbCl2 formed
= 278 * 6.5/223 * 278
= 0.029 moles
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❤BE BRAINLY ❤
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Dear student,
Given :-
- weight of PbO = 6.5 gm
- weight of HCl = 3.2 gm
To find :-
- moles of lead chloride PbCl₂ in the reaction
Solution :-
The given condition in the question can be stated in terms of chemical reaction as;
1PbO + 2HCl ⇒ 1PbCl₂ + 1H₂O
Using the concept of limiting reagent in the above given reaction, we have;
nPbO = w/ mw
where, n = moles
w = given mass
mw = molecular mass
= 6.5 / 223
&, nHCl = w/ mw
= 3.2 / 36.5
Then,
1 PbO 2 HCl
6.5/ 1×223 3.2 /2×36.5
6.5/ 223 3.2/73
0.029 0.043
From the above, clearly, 1 PbO is the limiting reagent. So, multiplying the co-efficient of PbCl₂ with the value of limiting reagent, we obtain;
0.029 × 1 ( PbCl₂ )
0.029 PbCl₂
Hence, 0.029 PbCl₂ moles of lead chloride will be formed.