Methanol,   CH3OH,   can be made by combining gaseous carbon monoxide and hydrogen gas.?
814744
In the presence of the proper catalyst:\n\nCO + 2 H2  -->  CH3OH\n\n1.40 g of H2 = 1.40 g / 2.016 g/mol = 0.694 moles H2\n\n13.85 g CO = 13.85 g / 28.01 g/mol = 0.4945 moles of CO\n\nIt takes 2 mole of H2 for every 1 mole of CO, so only 0.347 moles of CO will react.  This will give 0.347 moles of CH3OH:\n\n32.04 g/mol * 0.347 moles = 11.12 grams\n\n11.12 is your answer.
652303
Unfortunately this is not possible.\n\nWater is Hydrogen Dioxide (H20) whereas Methanol and Ethanol have Carbon in their chemical structure.  To change Water into Methanol or Ethanol you would first have to expend Enegy into cracking Water into its parts (Hydrogen and Oxygen) and then expend more energy adding Carbon to the Hydrogen in order to sythensis Methanol or Ethanol.
213015
NH3  +  CH3COOH = CH3COO–  +  NH4+\n\nThe gas would be NH4?
438289
Co(NH3)4 would be more likely
962530
isopropyl alcohol, CH3CHOHCH3
552992
ch4 for both.\nOne carbon atom and 4 hydrogen atoms. Simple!
978914
Chemically combined means that the two substances have reacted together to give a new substance, as opposed to just mixing.  Like, if you mixed hydrogen and oxygen together without a reaction, you'd still have hydrogen gas and oxygen gas.  But if they react together, they're chemically combined to give water.\n\nThis really should have been in chemistry rather than mathematics.
771136
NaC2H3O2 + H2O\n\nyou can't combine things unless they match with numerals. such as... you cannot add H3 and H because they are different chemicals/elements. if it were 3H + H you could combine them because its telling you that you have 3 hydrogens + and additional hydrogen. thus leaving you with NaC2H3O2 + whatever is left over. which would be an O and 2 H's. giving you the +H2O\n\nhope this helps!
807508
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
81567
Yes, you can make methane gas from any decomposing organinc material.  With methane gas you can generate power, fuel cars, etc.  We should be making lots of it.
1050074
mathanol. a research in the MIT found out a solution for saving fuel. According to them it increases the efficiency by 250 percent than conventional vehicle. Also the engine size will be reduced maintaining the perfomance. also the knocking will be reduced cuz methanol is more powerful than petrol.\nHydrogen will not be an alternative cuz it is dangerous to use it as a fuel. it wouldn;t be in use in vehicles unless and untill a safe unit is not developed.
55863
NH4(OH)\n\nIt is from combination of NH4+ , OH-
968613
If you want to make 1 Liter of this final solution, you would need to take 721 mL of your 52% formaldehyde solution, add 120 mL of methanol to it and bring the final volume up to 1 Liter with water.  You should be able to scale this up or down to make any particular volume you need.\n\nWhat isn't coming out right in your calculations?
453848
A reformer is a huge furnace that provided the heat enery required to seperate the Hydrogen and Carbon from the stable CH4 molecule. \n\nThe ones I have worked with works like this:\nA mixture of the gas CH4 and steam H2O is passed thorough a tube that is filled with catalyst, that goes into the furnace (reformer) where it is heated to around 900C.\n\nAt that temp, and with the steam, the following are formed\nHydrogen, Carbon Monoxide, and Carbon Dioxide. So the carbon forms the CO and CO2.\n\nOther molecules can cause the following: Oxygen: Explosion\nNitrogen: Nothing\nSulpur: Kills catalyst.
716080
fist of all yes a hydrocarbon is made up on H & C only\n Fist divid 5.91g of CO2 by it's RMM (44) =0.1343moles the 4.83 g of H2O by it's RMM (18) =0.2683\nthere for for every 0.1343moles of Carbon there is 0.2683moles of Hydrogen\n C :\n1/0.1343 =7.446x 0.2683(moles of H) = 1.997 (not 2 exactly because to the nearest no. was used)\nC:H\n1:2\ntherefore empirical formula CH2
963470
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour 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.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
900980
A 1,2-diol would form.\n\nEthene would form ethan-1,2-diol, (CH2OH)2.