We know,

1 mole of `Cl_2`=71 gm=22.4 l at STP

Thus, density,`rho=M/V=71/22.4=3.16` g/l

The required reaction is,

`CaCO_3+2HCl->CaCl_2+H_2O+CO_2`

Number of mole of HCl`=0.92**16.6/1000=0.015272\ mol e`

100 gm of `CaCO_3` reacts with 2 mole of HCl.

or, 1 gm of `CaCO_3` eact with 0.02 mole of HCl.

Thus, HCl is a limiting reagent.

We know, Pure `CaCO_3` in sample is that portion of `CaCO_3` which reacts completely.

Thus, % pure `CaCO_3=0.7636/1**100%`

`=76.36%`

We Know,

`(eqv.wt.of\ met al\ chlo ride)/(eq.wt.of\ me tal\ o x ide)=(wt.of\ met al\ chlo ride)/(wt.of\ me tal\ o x ide)`

let, equivalent weight of metal=x.

or,`(x+35.5)/(x+8)=5/3`

or,`x=(3**35.5-5**8)/(5-3)`

or,`x=33.25`

We Know,

1 molecule of `NH_3=` 7 neutrons

Now,

1 mole of `NH_3=22.4l=N_A\ NH_3` molecules

or, 2.24l of `NH_3=N_A/10\ NH_3` molecules

`=N_A/10**7` neutrons

`=4.2161**10^23`

We Know,

let equivalent weight of metal`=x`

`(eq.wt.of\ acid)/(eq.wt.of\ Mg\ sal t)=(wt.of\ acid)/(wt.of\ Mg\ sal t)`
or,`(x+1)/(12+x)=2/2.602`
or,`x=35.5`
Now, equivalent weight of acid `=1+35.5=36.5`

Mol. wt of metal `=2**12=24`

Mol. wt of metal oxide `=24+16=40`

The required reaction is,

`Ca(OH)_2+CO_2->CaCO_3+H_2O`

From above balanced equation, We get,

1 mole of `Ca(OH)_2` react to give 1 mole of `CaCO_3`

or, 74 gm of `Ca(OH)_2` react to give 100 gm of `CaCO_3`

or, 3.7 gm of `Ca(OH)_2` react to give `100/74**3.7=5` gm of `CaCO_3`

Given,

Amount of `CaCO_3=90%\ of 20g=18g`

Amount of `HCl=36.5%\ of 50g=18.25g`

The required reaction is,

`CaCO_3+2HCl->CaCl_2+H_2O+CO_2` 

From above balanced reaction, we get,

100 gm of `CaCO_3` reacts with 73 gm of `HCl`.

or, 18 gm of  `CaCO_3` should reacts with 13.74 gm of `HCl`.

Thus, excess `HCl=18.25-13.74=5.11gm`

Now,

100 gm of `CaCO_3` reacts completely to give 56 gm of `CO_2`.

18 gm of `CaCO_3` reacts completely to give `56/100**18=10.08` gm of `CO_2`.

We know,

56 gm of `CO_2=22.4l` at STP

or, 10.0 gm of `CO_2=22.4/56**10.08=4.032l`

We Know,

1 mole of `H_2O=18g m=6.023**10^23` molecules

Mass of water in 1 ml=`rho**V=1**1=1 `g/cc

Now,

18 gm of water has `6.023**10^23` `H_2O` molecules.

1 gm water has `(6.023**10^23)/18=3.34**10^23` `H_2O` Molecules

Given,

weight of sodium =0.46 gm

weight of sodium salt =1.64 gm

equivalent weight of sodium =23

equivalent weight of acid =?

We Know,

`E_(Na)/E_(Na.sal t)=W_(Na)/W_(Na.sal t)`

or,`23/(23+x)=0.46/1.64`

or,`x=59 gm`

Now, equivalent weight of an acid,HX `=59+1=60`