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vaieri [72.5K]
3 years ago
10

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

ater . Suppose 91. g of sulfuric acid is mixed with 116. g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to significant digits.
Chemistry
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

maximum mass of sodium sulphate=132g

Explanation:

firstly balance the chemical reaction,

H_2 SO_4_aq+2NaOH_s=Na_2 SO_4_aq+2H_2 O_l

molecular weight of H2SO4=98g/mole;

Molecular weight of NaOH=40g/mole;

number of mole of H2SO4 given=91g/(98g/mole)=0.93mole;

number of mole of NaOH given=116g/(40g/mole)=2.9mole

moles are:

number of mole of H2SO4=0.93

number of mole of NaOH=2.9

From the balanced equation,

1mole of H2SO4 reacts with 2 moles of NaOH;

hence ;

0.93 mole of H2SO4 will react with 2*0.93 moles of NaOH means

moles of NaOH that reacts wih H2SO4=1.86mole but we have given 2.9 mole therefore

NaOH will be excess reagent and H2SO4 will be the limitting reagent

and mass of product depends on limitting reagent i.e mass of H2SO4.

1mole of H2SO4 produces 1 mole of Na2SO4;

0.93 mole produces 0.93 mole of Na2SO4;

maximum mass of Na2SO4 produces=mole*molecular weight;

Molecular weight of Na2SO4=142g/mole;

maximum mass of Na2SO4 produces=0.93*142=132g;

maximum mass ofNa2SO4 produces=132g;

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<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

                        = 80,000 g

Therefore;

Moles of water = \frac{80,000g}{18.02 g/mol}

                = 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O  available = 65.0 kg or 65,000 g

Molar mass Li₂O  = 29.88 g/mol

Moles of Li₂O  = 65,000 g ÷ 29.88 g/mol

          = 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

  • Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
  • However, the number of moles of water to be removed is 4.44 × 10³ moles  but only 2.175 × 10³ moles will react with the available Li₂O.
  • This means, Li₂O  is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water =  4.44 × 10³ moles  - 2.175 × 10³ moles

                                           = 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

                                          = 4.0815 × 10⁴ g or

                                          = 40.815 kg

Thus, the mass of excessive water is 40.815 kg

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