00:01
Hello everyone, balance the following equation.
00:03
Let us see the following question.
00:05
So, in option a, we are having the question where for the reaction to produce superphosphate.
00:12
So we have calcium phosphate, ca3, po4, and the whole twice, should be treated with sulfuric acid, h2o -s -4 in a ques form.
00:26
Then it forms calcium, hydrogen phosphat, speed h.
00:31
P .o .4, h2po4, the whole twice has to be treated with calcium sulfate.
00:43
Let us write the states.
00:45
So, this is solid.
00:48
This is acquies.
00:50
This is acquies.
00:51
This is acquies.
00:54
And also the other one is equious.
00:57
Okay, let us write the balanced equation.
00:59
Now that we know there are three calcium atoms.
01:03
So, there are only two atoms here.
01:05
We have to multiply anywhere two.
01:08
So that calcium, sulphate, suppose if i multiply two, this is tri -heaten trial method.
01:14
This sulfate becomes two.
01:16
So i need to multiply two here.
01:18
Hydrogen is totally four.
01:20
Already there is four units here.
01:23
Now calcium is 1 plus 2, 3 unit, sulfase 2 unit, hydrogen 4.
01:29
And phosphate is anyway two units.
01:32
So, this is the balanced equation in the first case.
01:35
Let us see the second case.
01:38
Nabh4, sodium boron, hydride, should be treated with sulfuric acid h2 -s -o -4 plus, sorry, it is giving diburin, that is b2 h6 plus hydrogen gas, h2, plus hydrogen gas, n -a -2 -s -o -4.
02:02
Let us write the states of reaction.
02:05
This is solid.
02:06
This is aqueous...