- <u>The reaction that takes place is:</u>
Hg(NO₃)₂(ac) + Na₂S(ac) → HgS(s) + 2Na⁺ + 2NO₃⁻
Now we calculate the moles of each reagent -using the molecular weights-, in order to determine the limiting reactant:
- Moles of mercury (II) nitrate = 85.14 g *
=0.2622 moles. - Moles of sodium sulfide = 14.334 g *
=0.1837 moles.
Because the stoichiometric ratio between the reactants is 1:1, we compare the number of moles of each one upfront.
moles Hg(NO₃)₂ > moles Na₂S
<u>Thus Na₂S is the limiting reagent.</u>
So in order to find the mass of solid precipitate, we must calculate it using the moles of Na₂S:
The mass of the solid precipitate is 42.760 g.
- In order to calculate the grams of the reactant in excess that will remain after the reaction, we convert the moles that reacted into mass and substract them from the original mass:
Mass of Hg(NO₃)₂ remaining = 
The mass of the remaning reactant in excess is 25.49 g.
- Because we assume complete precipitation, there are no more Hg⁺² or S⁻² ions in solution. The moles of NO₃⁻ and Na⁺ in solution remain the same during the reaction, so the number is calculated from the number added in the reactant:
Hg⁺²: 0 mol
NO₃⁻: 
Na⁺: 
S²⁻: 0 mol
Answer:
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Explanation:
Since sulfuric acid, H₂SO₄, is a diprotic acid and potassum hydroxide, KOH, contains one OH⁻ in the formula, the number of moles of potassium hydroxide must be twice the number of moles of sulfuric acid.
<u>1. Determine the number of moles of KOH in 47mL of 0.39M potassium hydroxide solution</u>
- number of moles = molarity × volume in liters
- number of moles = 0.39M × 47mL × 1liter/1,000 mL = 0.1833mol
<u>2. Determine the number of moles of sulfuric acid needed</u>
- number of moles of H₂SO₄ = number of moles of KOH/2 = 0.1833/2 = 0.009165mol
<u>3. Determine the concentration that contains 0.009165 mol in 25mL of the acid.</u>
- Molarity = number of moles / volume in liters
- M = 0.009165mol/(25mL) × (1,000mL/liter) = 0.3666M
Round to two significant figures: 0.37M
Step 7- Communicate. Present/share your results. Replicate.
Step 1- Question.
Step 2-Research.
Step 3-Hypothesis.
Step 4-Experiment.
Step 5-Observations.
Step 6-Results/Conclusion.
Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ
Explanation:
Using this formular, q = [mCpΔT] and = [nΔHfusion]
The energy that is needed in the different physical changes is thus:
The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:
q = [mCpΔT]
q = 52.0 x 2.09 x 10
q = 1.09 kJ
While from 0°C to 100°C is calculated as:
q = [mCpΔT]
q = 52.0 x 4.18 x 100
q = 21.74 kJ
And for fusion at 0°C is called Heat of fusion and would be given as:
q = n ΔHfusion
q = 52.0 / 18.02 x 6.02
q = 17.38 kJ
And that required for vaporization at 100°C is called Heat of vaporization and it's given as:
q = n ΔHvaporization
q = 52.0 / 18.02 x 40.7
q = 117.45 kJ
Add up all the energy gives 157.8 kJ
The reaction creates dihydrogen, hence if it's uncontrolled it could lead to potentially dangerous amounts of gas being released at once.
