Solution containing 43.20g mercury(II)perchlorate is allowed to react completely with a solution containing 7.410g sodium sulfide,how many grams of solid precipitate will form?
How many grams of the reactant in excess will remain after the reaction?
Let's write molecular formula
Mol wt=200.592+2(35.5+64)=399.592 out of which Hg is 200
If all the mercury combines with sulphide the formula weight of mercury sulphide is 200.592+32=232.592
What we have is 43.2 gm of a compound of mol wt 399.592(mercury perchlorate) resulting after complete reaction into formula wt of 232.592(mercury sulphide)
So our product to reactant ratio is 232.592/399.592
If we multiply it with actual weight of provided mercury perchlorate we should be getting product of mercury sulphide as 43.2x232.592/399.592=25.1456gm
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