If a solution containing 80.701 g80.701 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 12.026 g12.026 g of sodium sulfide, how many grams of solid precipitate will form? precipitate: gg how many grams of the reactant in excess will remain after the reaction? excess reactant:

The excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.

Explanation:

To calculate the number of moles, we use the equation:

  ....(1)

For sodium sulfide:

Given mass of sodium sulfide = 12.026 g

Molar mass of sodium sulfide = 78.045 g/mol

Putting values in above equation, we get:

For mercury (II) perchlorate:

Given mass of mercury (II) perchlorate = 80.701 g

Molar mass of mercury (II) perchlorate = 399.49 g/mol

Putting values in above equation, we get:

For the given chemical equation:

Here, the solid precipitate is mercury sulfide.

By Stoichiometry of the reaction:

1 mole of sodium sulfide reacts with 1 mole of mercury (II) perchlorate.

So, 0.154 moles of sodium sulfide will react with = of mercury (II) perchlorate

As, given amount of mercury (II) perchlorate is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfide is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (mercury (II) perchlorate) left = 0.202 - 0.154 = 0.048 moles

By Stoichiometry of the reaction:

1 mole of sodium sulfide reacts with 1 mole of mercury sulfide.

So, 0.154 moles of sodium sulfide will react with = of mercury sulfide.

Now, calculating the mass of mercury sulfide from equation 1, we get:

Molar mass of mercury sulfide = 232.66 g/mol

Moles of mercury sulfide = 0.154 moles

Putting values in equation 1, we get:

Hence, the excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.

Hope the image of the frequency formula

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.