- Messages
- 107
- Reaction score
- 32
- Points
- 38
Q. 2.80g of a group II metal carbonate, XCO3, is added to 50cm^3 of 2.00 moldm^-3 HCl (excess) in a plastic cup. the temperature rose by 9.7K. Calculatre the Mr of the carbonate.
for this question, can we assume that q= sum of all the bond energies of bonds involved in the formula q=mcdT? I cant think of any other technique..
XCO3 + 2HCl ---> XCl2 + H2O + CO2
2.80g of XCO3 (ignore it for now)
x mol of XCO3 react with 2x mol of HCl
Moles of HCl = Volume(cm^3) multiplied by concentration(mol dm-3) DIVIDED by 1000
--> (50*2.00) / 1000
--> 0.1 mol
Compare the mole ratio.
If 2x = o.1 mol
then x = 0.05 mol
Now that you have the moles and the mass, find the Mr by using this:
Mr = mass/mol
--> 2.8/0.05
--> 56
Is this the correct answer?
Do forgive me if I wrote anything incorrect. Allah knows best.