00:01
So here, given that, volume of h2sf4 divided by volume of naaoh is 2 by 1, volume of nah is equal to 100 ml, volume of h2sf4 is equal to 200 m.
00:33
Now, molarity of nah is half into normality, which is equal to half into 10, it is 5 molar.
00:49
Now, moles of h2 sfo is 5 molar, this mole per liter into 200 by thousand, which is equal to 1 more.
01:05
Now, the molecular rate of h2 sfo is 98 gram per mole, that of naaoh is 40 gram per mole, fat of na2 s4 is 142 gram per mode.
01:35
The reaction is h2 s4 reacting with 2 moles of n .a .o .h giving n8 to s4 plus 2 moles of h2.
01:47
So in the first part of the question, it is asked to find the limiting reactor as well as the excess reactor.
01:53
So, naoh supplied is 100m, which is equal to 100 ml, multiplied by 1 .5 gram per ml, which is equal to 1 .5 gram per ml, which is equal to 1.
02:16
51 .5 gram.
02:22
Now mass of n .a .o .chin solution is 0 .55 multiplied by 151 .5 which is equal to 83 .3 .3 gram.
02:41
Moles of nahs is supplied is 2 .083 modes.
02:53
So, 135 gram n8 2 s4 is equal to 0 .95 more n8 2 .7.
03:02
So, from react 5 .5.
03:03
So, from react 5, 4.
03:03
So, from, station stociam entry for a 95 mole h2 s4 will react with for a 95 mole of n .a2 s4...