CH4 + H2O -> CH3OH + CH3OCH3
Does not balance;
Each compound on the product side has one Carbon and Hydrogen atom, so there must be one methane and water molecule per each DME and methanol molecule
So the hydrogens are then impossible to balance as one methane molecule has four hydrogens add that to the two hydrogens from each water molecule thats 6.
Methanol has four hydrogens - bearing in mind that 6 would be produced per molecule
The reaction hence cannot be balanced
2 CH3OH -> H2O + CH3OCH3
is the sythesis of DME from its consituents, maybe there was a transcription error somewhere.
3rd (methanol is polar, asym.O), then 1st (DMK is apolar, asym. O), then 2nd (gas)
Co + 2h2 -> ch3oh
CH3OH is molecular.
What is the balanced equation for Ch3COOH alone
Ch3oh
Ch3och3
Co + 2h2 -> ch3oh All letters capitals!
chcl3+CH3OH
3rd (methanol is polar, asym.O), then 1st (DMK is apolar, asym. O), then 2nd (gas)
As with any combustion reaction you need to include oxygen. The full equation for methanol combustion is: CH3OH + 2 O2 --> CO2 + 2 H2O
CuO + CH3OH --> HCHO + Cu + H2O
Co + 2h2 -> ch3oh
9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH
CH3OH is molecular.
9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH
What is the balanced equation for Ch3COOH alone
The formula of formaldehyde is CH2O. The products formed are sodium formate (HCOONa) and methanol (CH3OH). The stoichiometric equation is then X CH2O + Y NaOH --> A HCOONa + B CH3OH. Balancing the equation makes the coefficients 2 CH2O + NaOH --> HCOONa + CH3OH.