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When 1.50 g of Ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00°C to 33.10°C. If the specific heat of the solution is 4.18 J/(g ∙ °C), calculate for the reaction, as written. Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g)

Respuesta :

Answer:

The amount of heat is 431.12 kJ/mol.

Explanation:

Given that,

Mass of Ba = 1.50 g

Mass of water = 100.0 g

Initial temperature = 22.00°C

Final temperature = 33.10°C

The reaction is,

[tex]Ba+2H_{2}O\Rightarrow Ba(OH)_{2}+H_{2}[/tex]

We need  to calculate the heat

Using formula of heat

[tex]Q=ms\Delta T[/tex]

Where, m = mass

s = specific heat

[tex]\Delta T[/tex] temperature

Put the value into the formula

[tex]Q=(1.50+100)\times4.18\times(33.10-22.00)[/tex]

[tex]Q=4709.397\ J[/tex]

We need to calculate the amount of heat per mole

Using formula of energy per mole

[tex]per\ mole\ Q=\dfrac{4709.397}{moles\ of\ Ba}[/tex]

Put the value into the formula

[tex]per\ mole\ Q=\dfrac{4709.397}{\dfrac{1.5}{137.32}}[/tex]

[tex]per\ mole\ Q=431129.59\ J/mol[/tex]

[tex]per\ mole\ Q=431.12\ kJ/mol[/tex]

Hence, The amount of heat is 431.12 kJ/mol.

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