How many grams of CH2OH can be made from 36.0g of CO if it all reacts?
I guess we are looking at 2CO + 3H2 = 2CH2OH. So 1 mol CO yields one mol CH2OH. Mol wts. CO = 12+16 = 28 CH2OH = 12+ 2*1 + 16 +1 = 31 36 g CO so 36/28 mol and gives that number of moles of CH2OH which is 31*36/28 g = 39.9 g
How many grams of CH3OH can be made from 18.0 g of CO if it all reacts?
CO + 2H2 ===⇒ CH3OH 18.0 grams CO x 1mole CO/28 g x 1 mole CH3OH/1mole CO = 0.643 moles CH3OH 0.643 moles CH3OH x 32 grams CH3OH/1mole = 20.6 grams CH3OH Percent Yield: 14.0 g/20.6 g x 100 = 68.0%
If 4.0 of H2 are made to react with an excess of CO, according to the following equation: CO+H2---->CH3OH,...?
CO + 2 H2 → CH3OH Supposing the missing units in "4.0 of H2" to be "grams": A) (4.0 g H2) / (2.015894 g H2/mol) x (1/2) x (32.04200 g CH3OH/mol) = 32 g CH3OH in theory B) (28.0 g) / (32 g) = 0.88 = 88%
(a) How many grams of CH3OH can be made from 24.5 g of CO if it all reacts?
(a) CO(g) + 2 H2(g) -----> CH3OH(l) Molar mass of CO = 28 g/mol Molar mass of CH3OH = 32 g/mol According to equation; 1 mol CO which is 28 g produces 1 mol CH3OH which is 32 g. 24.5 g CO will produce; 24.5 x 32 / 28 = 28 g CH3OH will be produced theoretically (b) Percent yield = (actual yield / theoretical yield) x 100 % yield = (9.49 g / 28 g) x 100 = 33.9 %
Limiting Reactants - 10 Points!?
Methyl alchohol, CH3OH, is a clean-burning, easily handled fuel. It can be made by the direct reaction of CO and H2. Assume you start with 12.0 g of H2 and 74.5 g of CO. CO(g) + 2 H2(g) CH3OH(l) a. Which of the reactant(s) is the limiting reactant? b. Which of the reactant(s) is in excess? c. What mass of the excess reactant is left after the reaction is complete? d. How many grams of methyl alcohol can be obtained theoretically? a. We need to take the masses for each reactant and convert them both to product… b. We’ve already answered b, in that we label carbon monoxide as the limiting reactant therefore, H2 is in excess. c. We now need to use our LR CO and calculate the amount of H2 that is used during the reaction. d. We’ve already answered d too, in that our LR carbon monoxide will dictate the amount of product formed, 85.2 g. PERCENT YIELD We can take these limiting reactant calculations one step further and get a measure of the efficiency of the reaction itself. Any calculation using stoichiometry gives a theoretical value. A stoichiometric value of expected product would only exist in a perfect world with a 100% yield. However, in the real world, Example 6: When 16.7 g of calcium chloride is reacted with 18.2 g of sodium carbonate, 13.9 g of calcium carbonate is produced. CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2 NaCl(aq) We need to determine the Limiting reactant in order to determine the theoretical yield. So if all of the H2 is used up it would potentially produce 95.1g of product, while using up the entire mass of CO we could only expect 85.2g of product. The CO would run out before H2 therefore CO is our LR. 10.7 g of H2 is used during the reaction to produce the product. We started with 12.0 g therefore, there should be __________ left over.
How many grams of CH3OH will be produced?
1st write the balanced equation: 2 H2 + CO ---> CH3OH Now calculate the moles of hydrogen gas and carbon monoxide. Take the grams for each and divide them by their respective molar masses (found on periodic table). Now find the limiting reagent. It takes 2 moles of hydrogen gas to react with 1 mol of carbon monoxide. So if the amount of moles of hydrogen gas is more than twice the amount of carbon monoxide, then carbon monoxide is the limiting reagent. If the amount of moles of hydrogen gas is less than twice the amount of moles of carbon monoxide, then hydrogen gas is the limiting reagent. After finding the limiting reagent, use the amount of moles found for it to calculate the amount of moles of methanol that are produced. If the limiting reagent is carbon monoxide, the moles of methanol will be the same as that of carbon monoxide. If the limiting reagent is hydrogen gas, the amount of moles of methanol produced would be half of the amount of moles of hydrogen. Finally, with the moles of methanol found, multiply that number by the molar mass (g/mol) of methanol to find your final answer in grams of methanol.
CO(g) + 2H2(g) → CH3OH(l)???
To find number of moles, mass/molecular weight, Moles of CO = 33.3g / 28.01g/mol =1.19mols Moles of H2 = 9.59g / 2.016g/mol = 4.72mols ________CO_____+____ 2H2 ---------> CH3OH Mols____1.19__________ 4.72_______1.19 (The underscores don't mean anything, just want to illustrate the position of the numbers) H2 is the excess reactant since theoretically it only requires, 1.19 mols x 2 = 2.38mols (1 mole of CO react with 2 moles of H2) therefore 4.72 - 2.38 = 2.34g unreacted H2 Theoretical yield of alcohol = mols of CH3OH x molecular weight = 1.19mols x 32.042g/mol = 38.13g
What is the reaction between CaCl2 + H2O?
This is the reverse reaction of the acid - base neutralization reaction; 2HCl+Ca(OH)2 = CaCl2 + 2H2O. The Ca2+ and Cl- ions will stay in aqueous solution, and the backwards reaction of CaCl2 with H2O just won’t happen. That is, the chemical equilibrium of the above reaction is all the way over to the right. This is because the reverse reaction is energetically very unfavorable.The reverse reaction is equivalent to; H2O = H+ + OH -, that is, it's just the dissociation of water. This has an equilibrium constant, K(w) = 10^-14, at 25C, and so is negligible in solutions of pH 7.
Another chemistry homework question: "how many grams of X can be made from 40g of Y"?
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How many moles of hydrogen (H2) are needed to react with 8.0 g of O2?
2H2+O2------------->2H2OSo 4 g of Hydrogen react with 32 g of Oxygen to give 36 g of water.So 8 g Oxygen will react with=8 x4/32=2 g of HydrogenIn other words in the combustion of H2 with O2 they react in the ratio H2:O2:=1:8 .From this we come to know 8 g Oxygen will react with 1 g Hydrogen.!!!!