The amount of Al2O3 in moles= 1.11 moles while in grams = 113.22 grams
<em><u>calculation</u></em>
2 Al + Fe2O3 → 2Fe + Al2O3
step 1: find the moles of Al by use of <u><em>moles= mass/molar mass </em></u>formula
= 60.0/27= 2.22 moles
Step 2: use the mole ratio to determine the moles of Al2O3.
The mole ratio of Al : Al2O3 is 2: 1 therefore the moles of Al2O3= 2.22/2=1.11 moles
Step 3: finds the mass of Al2O3 by us of <u><em>mass= moles x molar mass</em></u><em> </em>formula.
The molar mass of Al2O3 = (2x27) +( 16 x3) = 102 g/mol
mass is therefore= 102 g/mol x 1.11= 113.22 grams
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
Salts is the correct awnser
