Asked by Jason
Liquid methanol (CH3OH) is combusted in oxygen gas to yield carbon dioxide gas and liquid water
Okay I keep getting this:
CH3OH(l) + 1.5 O2(g) > CO2(g) + 2 H2O(l)
the website tells me I'm wrong, but I don't know how to go about with out using fractions. Please help. Thanks
Okay I keep getting this:
CH3OH(l) + 1.5 O2(g) > CO2(g) + 2 H2O(l)
the website tells me I'm wrong, but I don't know how to go about with out using fractions. Please help. Thanks
Answers
Answered by
DrBob222
CH3OH(l) + 1.5 O2(g) > CO2(g) + 2 H2O(l)
You get rid of fractions in chemistry the same as in math. If it's a 1/2, just multiply by 2.
2CH3OH + 3O2 ==> 2CO2 + 4H2O
For 1/3 multiply by 3.For 1/4 multiply by 4, etc.
MOST times we need whole numbers to balance an equation. The only time that fractions are ok is in thermodynamic equations.
You get rid of fractions in chemistry the same as in math. If it's a 1/2, just multiply by 2.
2CH3OH + 3O2 ==> 2CO2 + 4H2O
For 1/3 multiply by 3.For 1/4 multiply by 4, etc.
MOST times we need whole numbers to balance an equation. The only time that fractions are ok is in thermodynamic equations.
Answered by
Jason
thank you sooo much
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