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Helga [31]
3 years ago
14

2al(s)+3cl2(g)→2alcl3(s) δh∘ = -1408.4 kj. how much heat (in kilojoules) is released on reaction of 4.70 g of al?

Chemistry
1 answer:
Annette [7]3 years ago
7 0
Answer:
             - 122.58 kJ

Solution:
According to following  equation,

                          <span>2 Al  +  3 Cl</span>₂    <span>→    2 AlCl</span>₃    <span> δH</span>°<span> = -1408.4 kJ
</span>
When,
         54 g (2 mole) Al on reaction releases  =  <span>1408.4 kJ heat
So,
                               4.70 g of Al will release  =  X kJ of Heat

Solving for X,
                      X  =  (4.70 g </span>× 1408.4 kJ) ÷ 54 g

                      X  =  - 122.58 kJ
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If you begin with 2.7 g Al and 4.05 g Cl2, what mass of AlCl3 can be produced?
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A 2.17 gm sample barium reacted completely with water what is the equation for the reaction how many milliliters of dry H2 evole
vaieri [72.5K]

Answer:

400 mL

Explanation:

Given data:

Mass of barium = 2.17 g

Pressure = 748 mmHg (748/760 = 0.98 atm)

Temperature = 21 °C ( 273+ 21 = 294k)

Milliliters of H₂ evolved = ?

Solution:

chemical equation:

Ba + 2H₂O →  Ba(OH)₂ + H₂

Number of moles of barium:

Number of moles = mass/ molar mass

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Number of moles = 0.016 mol

Now we  will compare the moles of barium with H₂.

                       Ba        :       H₂

                         1         :         1

                  0.016        :     0.016

Milliliters of H₂:

PV = nRT

V = nRT/P

V = 0.016 mol ×  0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm

V = 0.39 atm. L/0.98 atm

V = 0.4 L

L to mL

0.4 × 1000 = 400 mL

8 0
3 years ago
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