Question

1. The addition of 3.27 g of Ba(OH)₂ · 8 H₂O to a solution of 1.14 g of NH₄SCN in 175 g of water in a calorimeter caused the temperature to fall by 4.9°C. Assuming the specific heat of the solution and products is 4.20 J/g·°C, calculate the approximate amount of heat (in J) absorbed by the reaction, which can be represented by the following equation.

Ba(OH)₂ · 8 H₂O(s) + 2 NH₄SCN(aq) → Ba(SCN)₂(aq) + 2 NH₃(aq) + 10 H₂O(l)

-------- J

2. How many moles of butane, C₄H₁₀(l), must be burned to produce 210. kJ of heat under standard state conditions? (The heat of combustion for butane is −2855.7 kJ/mol.)

--------- mol

Answer 1

I can help u for the quetion 2

Given data : Heat of combustion of Butane (l) is -2855.7 kJ/mole.

I.e. when 1 mole of Butane is burnt there evolves 2855.7 kJ of heat energy.

Hence we write the equivalence equation as,

1 mole Butane= 2855.5 kJ heat energy

Now we are asked to find out the number of moles of butane to be burned to get 200 kJ of heat energy.

Hence we write,

If, 1 mole Butane = 2855.5 kJ heat energy

then say 'A' moles of Butane = 200 kJ heat energy.

On cross-muliplication we have,

A x 2855.7 = 1 x 210

A = 210 / 2855.7

A = 0.07354 = 7.354 x 10-2 moles.

0.07354 moles of Butane needed to be burned to get 210 kJ heat energy.

thank u