When 3.50 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 47.40°C. If the specific heat of the solution is 4.18 J/(g ⋅ °C), calculate ΔH for the reaction, as written. Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g)       ΔH = ?

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Asked Oct 30, 2019
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When 3.50 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 47.40°C. If the specific heat of the solution is 4.18 J/(g ⋅ °C), calculate ΔH for the reaction, as written.

Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g)       ΔH = ?

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Expert Answer

Step 1

Given

Mass of Ba(s) = 3.50g

Mass of water = 100.0g

Specific heat of solution = 4.18 J/g°C

Temperature rise = (47.40 – 22)°C

Step 2

The formula for the heat evolved

Q=mcAT
= 103.5gx4.18J/g°C(47.40-22) °C
=10988.802 J
10.9888 kJ
10.989 kJ
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Q=mcAT = 103.5gx4.18J/g°C(47.40-22) °C =10988.802 J 10.9888 kJ 10.989 kJ

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Step 3

This is the heat required to raise the temperature from 47.40 °C from 22.0°C

Therefore,

This is als...

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