Question

Question #32fe8

Answer 1

We're given an assumption and data to arrive at the conclusion. First we need to calculate the heat released from cooling water,

`q = mC_sDeltaT`

`q = 17.2g * (4.18J)/(g*°C) * (-80°C) * (kJ)/(10^3J) approx -5.75kJ`

of heat are released by cooling that water.

The equation for combusting butane is a combustion reaction,

`2C_4H_10 + 13O_2 to 8CO_2 + 10H_2O` where `DeltaH^° approx -5.75kJ`

This is where many will make a stupid mistake, there are 2 moles of butane in that reaction, so our calculations must take that into account when calculating the grams needed. I guarantee you there would be an answer on the test for this mistake.

`(-5.75kJ)/(2"mol") approx (-2.88kJ)/("mol")`

Moving on, we need,

`-350kJ * ("mol")/(-2.88kJ) * (58g)/("mol") * (kg)/(10^3g) approx 7.05kg`

of butane to release the quantity of heat desired.