00:01
In this problem, we're given 15 .62 grams of the specified species for a reaction.
00:09
I'm going to leave the states off for brevity.
00:14
And then what are we supposed to find product? okay, for our first one, we're given the following chemical reaction.
00:47
And here's our substance.
00:54
15 .62 grams of cl2.
00:57
I will need a molar mass tab open here.
01:13
That's 70 .91 grams of cl2 per mole of i have a one to two mole ratio for my product kcl to my chlorine and my molar mass of kcl 74 .55.
01:37
This will equal 15 .62 times two times 74 .55 divided by 70 .91.
01:55
And this will produce 32 .84 grams of kcl.
02:02
That is my first answer.
02:10
My second problem is let me get over here.
02:19
2k plus br yields 2kbr.
02:28
Here's my reactant in question.
02:41
This time i'll leave my molar mass of br2.
02:44
Dibromine 159 .81.
02:53
Again, i have a one to two mole ratio for my product and a specified reactant.
03:16
Kbr has a molar mass of 119 .00.
03:34
So here i'll take 15 .62 times two times 119 divided by 159 .81.
03:46
And here i get 23 .26 grams of kbr.
04:02
My third problem, c, that one's good.
04:14
I have 4cr plus 3o2 and produce 2cr2o3.
04:25
And here's my substance.
04:29
So i have 15 .62 grams of o2 times 32 .00 grams of o2 per mole of o2.
04:43
My mole ratios for this based on my balanced chemical equation will be three moles of oxygen for two moles of chromium three oxide.
04:54
And let me get to my molar mass tab once again for cr2o3...