(Original post by Armaros)
The enthalpy of combustion of a substance refers to the enthalpy change when one mole of a substance reacts completely with oxygen.
The question says that chromium reacts to produce chromium (III) oxide, and gives us the formula for this in the table: Cr2O3.
Writing out the balanced equation for chromium's combustion:
2Cr + 1.5O2 -> Cr2O3.
This is the same as the enthalpy of formation of Cr2O3, the value for which is given in the table as -1140. However, it is not the correct equation for the enthalpy of combustion of chromium, since 2 moles are involved instead of 1. So, to get the correct equation for chromium's enthalpy of combustion, we divide that equation by 2, and therefore also divide its enthalpy change (-1140) by 2, giving us the answer of -570 kJ/mol.
a) 2 moles thiosulphate react with 1 mole iodine
n = cv = 0.1 * 0.0205 = 0.00205 moles thiosulphate reacted
divide by 2 = 0.001025 moles iodine reacted
b) 1 mole ClO- produces 1 mole iodine so 0.001025 moles ClO- present in the 25cm3 of solution.
So 0.01025 moles present in the 250cm3 of solution.
A mistake here might be to calculate the concentration of ClO- in the solution, which would lead to 0.01025/0.25 = 0.041.
However, the question asks for the concentration of the ClO- in the bleach
, of which there is 10cm3 in the solution, so the correct equation is 0.01025/0.01 = 1.025 moles per litre.
From question 6 we know that 2 moles thiosulphate react with 1 mole iodine.
i) 0.025 * 0.0188 = 0.00047 moles thiosulphate reacted
so divide by 2 to get 0.000235 moles iodine
0.000235/0.01 = 0.0235 moles per litre
Same procedure for ii with different numbers!