499 grams of aqueous lead (II) nitrate reacts with potassium sulfide. Determine what products are formed, balance the chemical equation...?
2015-04-16 1:03 am
499 grams of aqueous lead (II) nitrate reacts with potassium sulfide. Determine what products are formed, balance the chemical equation and calculate the theoretical yield, in grams, of the product containing potassium
回答 (1)
2015-04-16 4:22 pm
Pb(NO3)2 + K2S → PbS + 2 KNO3
Supposing the K2S is in excess:
(499 g Pb(NO3)2) / (331.2098 g Pb(NO3)2/mol) x (2 mol KNO3 / 1 mol Pb(NO3)2) x
(101.10332 g KNO3/mol) = 305 g KNO3 in theory
Supposing the K2S is in excess:
(499 g Pb(NO3)2) / (331.2098 g Pb(NO3)2/mol) x (2 mol KNO3 / 1 mol Pb(NO3)2) x
(101.10332 g KNO3/mol) = 305 g KNO3 in theory
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