ile gramów CuO przereaguje z propan-2-ol aby powstało 100g ketonu
CH3-CH(OH)-CH3 + CuO --> CH3-C(=O)-CH3 + H2O + Cu
m CuO = 64g+16g = 80g
m C3H6O = 3*12g + 6*1g+16g = 58g
80g CuO ---- 58g ketonu
x -------------100g
x = 80g*100g/58g = 138g
Odp. Potrzeba 138g tlenku miedzi(II)
80u 58u
C₃H₇OH + CuO --> C₃H₆O + Cu + H₂O
x 100g
z proporcji wyliczam x
x ≈ 138g
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Litterarum radices amarae sunt, fructus iucundiores
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CH3-CH(OH)-CH3 + CuO --> CH3-C(=O)-CH3 + H2O + Cu
m CuO = 64g+16g = 80g
m C3H6O = 3*12g + 6*1g+16g = 58g
80g CuO ---- 58g ketonu
x -------------100g
x = 80g*100g/58g = 138g
Odp. Potrzeba 138g tlenku miedzi(II)
80u 58u
C₃H₇OH + CuO --> C₃H₆O + Cu + H₂O
x 100g
z proporcji wyliczam x
x ≈ 138g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)