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

The amount of heat is 431.12 kJ/mol.

Step-by-step 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)*4.18*(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=(4709.397)/(moles\ of\ Ba)`

Put the value into the formula

`per\ mole\ Q=(4709.397)/((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.