A student added 10g of 10%lead nitrate to 10 gram of 10% sodium chloride. after the reaction was complete, he found 18 gram of solution .what was the amount of lead chloride formed?(could someone please explain it's urgent I got my exam tomorrow!)
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Pb (NO3) 2 + 2NaCl--------> PbCl2 + 2 NaNO3.
Molecular Weight of Pb (NO3) 2 = 331 g / mol
Molecular weight of NaCl = 58.5 g / mol
So 10 g of Pb (NO3) 2 will react with
=10 x 58.5 / 331 g of NaCl = 1.75 g of NaCl.
The amount of PbCl2 formed = 10 + 1.75 = 11.75 g.
The rest of ( 10 – 1.75 ) = 8.25 g of NaCl, will remain unreacted.
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Hi! I hope this could help you in your exam.
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