you are ok (I didn't check math) until here:
For every 1 mole Hg(NO3)2, there are 2 moles Hg(NO3)2 ****That is nuts
Limiting Reagent: Hg(NO3)2***correct
Product = 475g****Nuts. You get .0269 moles of HgI2, that is the product. figure the mass of that.
Determine the type of chemical reaction, find the limiting reagent, determine the amount of product in grams, and the amount in grams of excess reagent.
8.75 g of mercury (II) nitrate solution is mixed with 9.83 g sodium iodide solution (ppt).
Hg(NO3)2(aq) + 2NaI(aq) --> HgI2(s) + 2NaNO3(aq)
Chemical reaction: double replacement
Molar mass Hg(NO3)2 = 325amu. 1moleHg(NO3)2 = 325gHg(NO3)2
(1moleHg(NO3)2 / 325gHg(NO3)2) 8.75gHg(NO3)2 = .0269molesHg(NO3)2
Molar mass NaI = 150amu. 1moleNaI = 150gNaI
(1moleNaI / 150gNaI) 9.83gNaI = .0655molesNaI
For every 1 mole Hg(NO3)2, there are 2 moles Hg(NO3)2
Limiting Reagent: Hg(NO3)2
Product = 475g
I am not sure if I am figuring out the number of moles or grams in this problem. Also, I am not quite sure how to determine the limiting reagent. Please check/correct my work so I can get a better understanding on how to do this. Thank you!
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