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Tema [17]
3 years ago
13

C,H,O,+ 3HCI → C,HCl, + 3H20 right 5 possible mole ratios

Chemistry
1 answer:
Marina86 [1]3 years ago
4 0

Answer:

Here's what I get  

Explanation:

I think this may be the equation you intended to write:

C₃H₅(OH)₃ + 3HCl ⟶ C₃H₅Cl₃ + 3H₂O

The mole ratios are the ratios of the coefficients in the balanced equation.

Here are some of the possible molar ratios.  

  1. C₃H₅(OH)₃:HCl        = 1:3
  2. C₃H₅(OH)₃:C₃H₅Cl₃ = 1:1
  3. C₃H₅(OH)₃:H₂O       = 1:3
  4.           HCl:C₃H₅Cl₃ = 3:1
  5.           HCl:H₂O       = 3:3
  6.    C₃H₅Cl₃:H₂O       = 1:3

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3 years ago
In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera
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Explanation :

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q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

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c_1 = specific heat of calorimeter = 1.55J/^oC

c_2 = specific heat of water = 4.184J/g^oC

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

q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]

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Now we have to calculate the enthalpy change of dissolution of KClO_4

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of KClO_4 = 2.84 g

Molar mass of KClO_4 = 138.55 g/mol

\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole

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Therefore, the enthalpy change of dissolution of KClO_4 is 54.3 kJ/mole

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