An impure Sample CaCO3 weighing 2.5g is completely reacted with 40ml of 0.5M H2So4.The Percentage of Purity of the Sample?

Question

callisto · Answer

$$CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + CO_2+H_2O$$
1. Find the no. of mole of H2SO4
2. No. of mole of CaCO3 reacted = no. of mole of H2SO4
3. Mass of CaCO3 reacted =No. of mole of CaCO3 reacted x molecular mass of CaCO3
4. % = answer in 3 / 2.5 x100%

PS: I haven't done chemistry for long

anonymous · Answer

Yup!

anonymous · Answer

$$H_2SO_4 + CaCO_3 --> CaSO_4 + H_2O + CO_2$$

It is balanced, I hope this is what you are looking for.

No. of moles = $$H_2SO_4=(1*2)+32+(4*16)= 2+32+64 = 98 mol$$

Applying the law of conservation of mass = 
 Mass of CaCO3 reacted =No. of mole of CaCO3 reacted x molecular mass of CaCO3

anonymous · Answer

Do you ge that @Yahoo!

yrelhan4 · Answer

@rohangrr what does percentage purity actualy mean?