Asked by michelle

What mas of liquid mercury should be produced by the reaction between 5.0 mL of 0.1102 M mercury(II)nitrate ans 0.621 grams of silver metal?

would the formula be...
Hg(NO3)2(aq) + 2Ag(aq)--> 2AgNO3(aq) + Hg(l)
Answered by DrBob222
Look at the limiting reagent problem you posted just above. This is worked the same way. Your equation looks ok.
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