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Arturiano [62]
3 years ago
13

f a solution containing 85.14 g of mercury(II) nitrate is allowed to react completely with a solution containing 14.334 g of sod

ium sulfide, how many grams of solid precipitate will be formed? mass: 29.69 g How many grams of the reactant in excess will remain after the reaction? mass: g Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. Hg2+ : 0 mol NO–3 : 0 mol Na+ : 0 mol S2− : 0 mol
Chemistry
1 answer:
Phantasy [73]3 years ago
6 0
  • <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 * \frac{1mol}{324.7g}=0.2622 moles.
  • Moles of sodium sulfide = 14.334 g *\frac{1mol}{78.04g}=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:

0.1837 molNa_{2} S*\frac{1molHgS }{1molNa_{2}S}*\frac{232.66g}{1molHgS} =42.740g

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 = 85.14g-(0.1837molHg(NO_{3})_{2} * 324.7 g/mol)=25.49g

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⁻: 0.2622molHg(NO_{3})_{2} *\frac{2molNO_{3}^{-}}{1molHg(NO_{3})_{2} *} =0.5244molNO_{3}^{-}

Na⁺: 0.1837molNa_{2} S*\frac{1molNa^{+}}{1molNa_{2}}=0.1837molNa^{+}

S²⁻: 0 mol

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A piece of rubber has a mass of 495 kg and a volume of 355 liters. Will it float in water, which has a density of 1.0 g/mL?
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Compare the density of the rubber to water. If it is less that 1 g/mL then it floats in the water.

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s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g
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This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express  your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol

The given chemical reaction is:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = \frac{3}{2}\times 0.05=0.0750mol of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Now put all the given values in above equation, we get:

743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L

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<h3>The answer is 5.0 g/mL</h3>

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<h3>5.0 g/mL</h3>

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