1. The addition of 3.27 g of Ba(OH) 2 · 8 H 2 O to a solution of 1.14 g of NH 4

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

Explanation / Answer

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

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