00:01
All right, to figure out the theoretical yield of methanol, we're first going to solve for the moles of carbon monoxide using pv equals nrt.
00:07
So n is going to be equal to p, which is 232, times your volume, which is 1 .50 liters, all over 62 .4, and then times the temperature, which is 305 kelvin.
00:28
Okay, so we're going to go ahead and round this mole off to three sigphags.
00:37
So this is 0 .5 .5 .5s.
00:37
So this is 0 .5 .5.
00:39
0 .947 moles of carbon monoxide.
00:44
Then we're gonna use the coefficients to get over to methanol.
00:47
So 1 is in front of methanol, and then we have a 1 in front of carbon monoxide, and then we're gonna multiply by the molar mass of methanol, which is 32 .04.
01:04
So 0 .947 times 32 .04 is gonna give us 30.
01:11
30 .03, or i'm sorry, 30 .3 grams of methanol from carbon monoxide...