Question

When 04 mole of MgCO3 is added to a 100 ml of 6 M HCI solution the following reaction occurs MgCo, -HO — MgCl2-002+H2O Select the correct statement(s) about the reaction. At the end of the reaction 0.2 moles of HCI is left behind At the end of the reaction 0,1 moles of MgCO3 is left behind Volume of CO2 produced at STP is 6.72 L Volume of CO2 produced at STP is 4.48 L​

Answer 1

Answer:

Explanation:

Sodium carbonate reacts with hydrochloric acid:

Na2CO3(s)+2HCl(aq)→2NaCl(aq)+CO2(g)+H2O(l)

We can find the number of moles of HCl before the reaction.

We then use the titration result to find the number of moles remaining after the reaction.

By subtracting the two we can get the number of moles of HCl which have reacted.

From the equation we can find the number of moles of Na2CO3.

From this we get the mass of Na2CO3.

We can then work out the percentage purity.

Concentration = moles of solute / volume of solution.

c=nv

∴n=c×v

∴ Initial moles of HCl=0.1280×50.001000=6.400×10−3

The acid remaining is titrated with NaOH:

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

nOH−=0.1220×30.101000=3.6722×10−3

Since they react in a 1:1 ratio the no. of moles of