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KatRina [158]
2 years ago
11

The reaction of hydrogen bromide(g) with chlorine(g) to form hydrogen chloride(g) and bromine(g) proceeds as follows: 2HBr(g) +

Cl2(g)2HCl(g) + Br2(g) When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ is evolved. Calculate the value of rH for the chemical equation given.
Chemistry
1 answer:
VladimirAG [237]2 years ago
8 0

Answer:

The enthalpy of reaction per mole of HBr for this reaction = ΔrH = -40.62 kJ/mole.

Explanation:

2HBr(g) + Cl2(g) → 2HCl(g) + Br2(g)

When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ of heat is evolved, calculate the value of ΔrH for the chemical reaction.

Note that ΔrH is the enthalpy per mole for the reaction.

Molar mass of HBr (g) = 80.91 g/mol.

Hence, 1 mole of HBr = 80.91 g

23.9 g of HBr led to the reaction giving off 12.0 kJ of heat

80.91 g of HBr will lead to the evolution of (80.91 × 12/23.9) = 40.62 kJ heat is given off.

Hence, 40.62 kJ of heat is given off per 80.91 g of HBr.

This directly translates to that 40.62 kJ of heat is given off per 1 mole of HBr

Hence, the heat given off per mole of HBr for this reaction is 40.62 kJ/mole.

But since the reaction liberates heat, it means the reaction is exothermic and the enthalpy change for the reaction (ΔHrxn) is negative.

Hence, ΔrH = -40.62 kJ/mole.

Hope this Helps!!!

You might be interested in
How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
The mass of an atom comes from _____. A. the nucleus B. the neutrons C. the electro​​​​ns D. the protons
kramer

Answer:

The answer is C

Explanation:

4 0
3 years ago
As temperature increases, density increases.<br><br> True or False
Alex73 [517]
Density increases with the temperature, true.
4 0
2 years ago
Read 2 more answers
Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802
Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

3 0
2 years ago
Please help Please help
alexdok [17]
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
5 0
1 year ago
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