What is the theoretical yield of CH3OH ? I.E., how many grams of CH3OH should be formed?

Question

Methanol, CH3OH, can be made by combining gaseous carbon monoxide and hydrogen gas.
In one experiment 2.00 g of H2 is mixed with 18.50 g of CO.


What is the theoretical yield of CH3OH ? I.E., how many grams of CH3OH should be formed?

Answer 1

CO + 2H2 ---> CH3OH

moles H2: 2.00g/ 1g/mol = 2.00 mol
moles CO: 18.50g / 28g/mol = 0.6607mol

your limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.

theoretical yield, therefore, is (1.321/2)*100 = 66.05%

based from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,

weight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams

Answer 2

First, find the limiting reagent: What is the starting material that is used up before the other, and so limits the product that can be formed.

2gH2 x 1 mol H2/2gH2 = 1 mol H2
18.50gCO x 1 mol CO/28g CO = 0.66 mol CO
The balanced equation is 2H2 + CO ===> CH3OH
So 0.66mol CO would need 2 x 0.66 = 1.32mol H2
There is not enough H2. So CO is the limiting reagent. Therefore:

18.50gCO x 1molCO/28gCO x 1molCH3OH/1molCO x 32gCH3OH/1mol CH3OH

Notice the cancelling out of all the units to give the units of the answer.

Answer 3

Co Ch3oh

Answer 4

(a million) write balanced equation (2) calculate moles CO and moles H2 (3) be sure proscribing reagent (4) calculate moles CH3OH in accordance with moles of proscribing reagent (5) calculate mass CH3OH (a million) 1CO + 2H2 ---> 1CH3OH (2) moles = wt/mw moles H2 = a million.40 two / 2 = 0.seventy one moles moles CO = 12.35/28 = .40 4 moles (3) proscribing reagent from balanced equation, a million CO reacts with 2 H2 so .40 4 CO reacts with .88 H2. on account that we've 0.seventy one H2, and we prefer 0.88 to thoroughly react with CO, H2 is the proscribing reagent (4) moles CH3OH .seventy one moles CO ---> .355 moles CH3OH from balanced equation (5) mass CH3OH moles = wt/mw wt = moles x mw = .355 moles x 32 g/mole = 11.4 g methanol notice that there are 3 sig figs....