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bulgar [2K]
3 years ago
14

What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?

Chemistry
1 answer:
Nataly [62]3 years ago
6 0

Complete Question

Magnesium sulfate forms a hydrate with the formula MgSO_4. 7H_20. What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of MgSO_4 is 120.4 g/mol; H20 = 18 g/mol.

Answer:

The value  is  z =  0.2093 \  g of  H_2O

Explanation:

From the question we are told that

   The volume of toluene is  V = 15 mL

    The mass of  anhydrous magnesium sulfate is  m =  200m g  = 200 *10^{-3} \  g

   The formula of the hydrate is   MgSO_4. 7H_20

    The molar mass of   MgSO_4  is  z =120.4 \ g/mol

From the formula given we see that

  1 mole of  Mg SO_4 wil remove  7 moles of H_2O to for the given formula

Hence

  120.4 g (1 mole) will remove  7 moles (7 * 18 g = 126 g  ) of  H_2O to for the given formula

Therefore 1 g of  Mg SO_4  x g  of  H_2O  

So

     x  =  \frac{x]126 *  1}{ 120.4 }

=>     x  =  1.0465 \  g

From our calculation we obtained that

  1 g of Mg SO_4 will remove  x  =  1.0465 \  g  of  H_2O  

Then  

   200 *10^{-3} \  g of Mg SO_4 will remove z g of  x  =  1.0465 \  g  of  H_2O  

So

   z =  200 *10^{-3} *  1.0465

=>z =  200 *10^{-3} *  1.0465

=>z =  0.2093 \  g of  H_2O  

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How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th
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Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

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We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

where,

ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

Gas: from 78.4°C to 135.3°C

Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

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