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rusak2 [61]
3 years ago
7

Using the balanced equation given below, calculate the number of moles of CaBr2 produced in the reaction of 5.0 moles of AlBr3.

____ moles CaBr2
*your answer should be written as X.X

2AlBr3 + 3CaO —> Al2O3 + 3CaBr2
Chemistry
1 answer:
allochka39001 [22]3 years ago
4 0

Answer:

7.5 moles of CaBr2 are produced

Explanation:

Based on the equation:

2AlBr3 + 3CaO → Al2O3 + 3CaBr2

<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>

<em />

Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:

5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =

<h3>7.5 moles of CaBr2 are produced</h3>

<em />

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K    52.10     52.10/39.10 = 1.33         1.33/1.32 ≈ 1
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O     32.1      32.1 / 16       =  2.01        2.01/1.32 ≈ 1.5

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
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Answer : The enthalpy of neutralization is, 56.012 kJ/mole

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\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

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From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

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q = heat absorbed = ?

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m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

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

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

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Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

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

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n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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