hey. my teacher barely taught us anything before break andthen send us home with

Question

hey. my teacher barely taught us anything before break andthen send us home with this giant packet of questions. so heresnumber one. i definately need help:

A solution containing 2.00 g of Hg(NO3)2 wasadded to a solution containing 2.00 g of NA2S. Calculatethe mass of products formed, the limiting reagent, and the excessreagent according to the reaction
Hg(NO3)2(aq) + Na2S(aq) --->HgS(s) + NaNO3(aq)

pleaseee help.

Explanation / Answer

2g of Hg(NO3)2 = 2/324.59 = 0.001616 moles 2g of Na2S = 2/78 =0.0256 moles => since moles of Hg(NO3)2 is less than moles ofNa2S => Hg(NO3)2 will act as Limiting reagentand NA2S is the excess reagent Hg(NO3)2 + NA2S -------->HgS(s) + NaNO3(aq) 0.001616     0.0256                   0         0 (intial numberof moles) 0       0.023984           0.01616 0.01616 (moles at equilibrium) So at equilibrium   1.87 gm of Na2S , 3.758gm of HgS , 1.373 gm of NaNO3 will be formed

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