Question

Calculate the mass of lead chloride formed by treating an aqueous solution 13.24 g of lead nitrate
with excess of hydrochloric acid
(Pb = 207. = 35.5, H = 1,O = 16)
​

Answer 1

Answer:

Mass of lead chloride formed = 5.56 g

Explanation:

207 + (14 +16 x 3) x 2 207 + (35.5 x 2)

207 + (14 + 48) x 2 207 + 71

207 + 62 x 2 278

207 + 124

331 g 278 g

331 g of lead nitrate yield 278 g of lead chloride. Hence, mass of lead chloride formed from 6.62 g of lead nitrate is -

Pb(NO3)2 : PbCl2

331 : 278

6.62 : x

Mass of lead chloride formed = 5.56 g

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