2 CH3-CH2-OH + 2 Na = 2 CH3-CH2-ONa + H2
m( C2H5OH) =100g
Mr(C2H5OH)=46g/mol
n=m/Mr = 100/46=2.17mol
2mol:1mol=2.17:x
x=2.17/2=1.08mol H2 but, this is not right, and don't know how to use m(Na)
V(H2)= Vm*n=22.4* number of moles of H2
I don't know how to find n(H2)
Yes, I've just solved it, thank you anyway. But, I need help with another problem:
What is the volume of H2 (normal conditions) which gets released when we add 4.6g of elementary Na into 100g of pure ethanol?
How many primary alcohols, which 3.7g releases 560 cm3 H2 in reaction with Na, are there?
I tried to find Mr of alcohol, but I can't get the right solution.
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