Chemistry

Hello everyone, just wondering if someone could check the question below for me. Much appreciated.

Vinegar can be used to neutralise magnesite, MgCO3, an ingredient in the table salt used for its anti-cacking properties. However in excess magnesite can have strong laxative properties. The reaction between the ethanoic acid in vinegar and magnesite can be represented by the following equation:
CH3COOH + MgCO3 = (CH3COO)2Mg + H2O + CO2
Morton Salt, Chicago, USA has 12 tonnes of salt containing 2% magnesite, they want to neutralise. Calculate the volume of your undiluted vinegar that would be required to neutralise all the magnesite in the slat.
My work out:
12 tonnes × 1000000/(1 tonne) = 12000000 g
2% of magnesite = (2/100) × 12 000 000 = 20 000 g
Mole of MgCO3 = (20000 g)/(84.3 g/mol) = 237.25 mol (Mr of MgCo3 = 84.3 g/mol)
Mole of CH3COOH = 237.25 × 2 = 474.50 mole
Volume of CH3COOH = (474.50 mol)/(1.0328 mol/dm3) = 459.43 dm3
Or
459.43 dm3 × (1000 cm3)/(1 dm3) = 4.59430 × 10^5 cm^3

Reply 1

whatacrydonnie

2% of 12,000,000 is not 20,000

Reply 2

sucik

ohhh yh my bad. its 240000g
is the rest okay?

Reply 3

whatacrydonnie

you'll need to redo all your calculations because the no of moles of MgCO3 will be 240,000/84.3
where did you get the concentration for CH3COOH from?

Reply 4

sucik

The concentration of CH3COOH = 1.0328 mol/dm3. I worked that out in previous question, where the mole was = 2.582 × 10^-2mol and volume was = 0.025 dm³.

12 × 1000000 = 1.2 × 10^6 g
2% of MgCO3 = 0.02 × 1.2 × 10^6 g = 2.4 × 10^5 g
Mol of MgCo3 = 2.4 × 10^5 g/84.3 g/mol = 2.847 × 10^3 mol
Mole of CH3COOH = 2.847 × 10^3 mol × 2 = 5.694 × 10^3 mol
Volume of CH3COOH = 5.694 × 10^3 mol / 1.0328 mol/dm3 = 5.513 × 10^3 dm3
Or 5.513 × 10^3 dm3= 5.513 × 10^3 cm3

Thank you