Koliko se molekula vode adira na propen ako se iz tako dobijenog proizvoda u reakciji sa dovoljnom količinom natrijuma oslobodi 56 cm3 gasa ? ( RESENJE 3*1021)
CH3CH=CH2+H2O---------->CH3CH2CH2OH
2CH3CH2CH2OH+2Na------------->2CH3CH2CH2ONa+H2
1 mol---------22.4dm3
x mol--------0.056dm3
x=0.056*1/22.4=0.0025 mol H2
n(H2):n(CH3CH2CH2OH)=1:2
n(CH3CH2CH2OH)=2*n(H2)/1=2*0.0025/1=0.005 mol
n(vode):n(CH3CH2CH2OH)=1:1
n(vode)=n(CH3CH2CH2OH)=0.005 mol
1 mol---------6*10(23)
0.005 mol---x
x=0.005*6*10(23)/1=3*10(21) molekula vode
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broj molekula vode
CH3CH=CH2+H2O---------->CH3CH2CH2OH
2CH3CH2CH2OH+2Na------------->2CH3CH2CH2ONa+H2
1 mol---------22.4dm3
x mol--------0.056dm3
x=0.056*1/22.4=0.0025 mol H2
n(H2):n(CH3CH2CH2OH)=1:2
n(CH3CH2CH2OH)=2*n(H2)/1=2*0.0025/1=0.005 mol
n(vode):n(CH3CH2CH2OH)=1:1
n(vode)=n(CH3CH2CH2OH)=0.005 mol
1 mol---------6*10(23)
0.005 mol---x
x=0.005*6*10(23)/1=3*10(21) molekula vode