Question

A solution containing 2.0 grams of Hg(NO3)2 was added to a solution containing 2.0 grams of Na2S. Calculate the mass of the HgS that was formed ...

Answer 1

To calculate the mass of HgS formed when 2.0 grams of Hg(NO3)2 reacts with 2.0 grams of Na2S, we need to determine the limiting reactant first. This can be done by calculating the number of moles of each reactant using their molar masses. The mass of HgS formed when 2.0 grams of Hg(NO3)2 reacts with 2.0 grams of Na2S is 2.32 grams.

Step-by-step explanation:

The balanced equation for the reaction is:

Hg(NO3)2 + Na2S -> HgS + 2NaNO3

Let's calculate the number of moles of Hg(NO3)2 and Na2S:

1. Number of moles of Hg(NO3)2 = mass / molar mass
2. = 2.0 g / 200.59 g/mol
3. = 0.00998 mol
4. Number of moles of Na2S = mass / molar mass
5. = 2.0 g / 78.11 g/mo
6. l = 0.0256 mol

From the balanced equation, we can see that the mole ratio between Hg(NO3)2 and HgS is 1:1.

Therefore, the mass of HgS formed can be calculated as:

Mass of HgS = number of moles of HgS * molar mass of HgS

= 0.00998 mol * 232.67 g/mol

= 2.32 g