Question

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)

Answer 1

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,

$Ba+2H_{2}O\Rightarrow Ba(OH)_{2}+H_{2}$

We need  to calculate the heat

Using formula of heat

$Q=ms\Delta T$

Where, m = mass

s = specific heat

$\Delta T$ temperature

Put the value into the formula

$Q=(1.50+100)\times4.18\times(33.10-22.00)$

$Q=4709.397\ J$

We need to calculate the amount of heat per mole

Using formula of energy per mole

$per\ mole\ Q=\dfrac{4709.397}{moles\ of\ Ba}$

Put the value into the formula

$per\ mole\ Q=\dfrac{4709.397}{\dfrac{1.5}{137.32}}$

$per\ mole\ Q=431129.59\ J/mol$

$per\ mole\ Q=431.12\ kJ/mol$

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