CHEMIE : Stoichiometrie oefeningen pag. 48
CHEMIE : Stoichiometrie oefeningen pag. 48
CHEMIE : Stoichiometrie oefeningen pag. 48
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56. Bereken hoeveel gram zuurstofgas nodig is om 20 g glucose te verbranden.<br />
Reactie: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O<br />
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O<br />
n = 1 mol n = 6 mol n = 6 mol n = 6 mol<br />
m = 20 g<br />
n = m / M<br />
= 20 g / 180 g/mol<br />
= 0,111 mol<br />
n = 0,667 mol n = 0,667 mol n = 0,667 mol<br />
m = n . M<br />
= 0,667 mol . 32 g/mol<br />
= 21,3 g<br />
57. Hoeveel gram dizuurstof is er nodig om 100 g ijzer om te zetten in Fe2O3?<br />
Reactie: 4 Fe + 3 O2 → 2 Fe2O3<br />
4 Fe + 3 O2 → 2 Fe2O3<br />
n = 4 mol n = 3 mol n = 2 mol<br />
m = 100 g<br />
n = m / M<br />
= 100 g / 55,8 g/mol<br />
= 1,79 mol<br />
n = 1,34 mol n = 0,90 mol<br />
m = n . M<br />
= 1,34 mol . 32 g/mol<br />
= 43 g<br />
58. Hoeveel kg vloeibare zuurstof is er nodig om 1 kg butaan (C4H10) volledig te verbranden tot CO2 en<br />
H2O? Reactie: 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O.<br />
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O<br />
n = 2 mol n = 13 mol n = 8 mol n = 10 mol<br />
m = 1000 g<br />
n = m / M<br />
= 1000 g / 58 g/mol<br />
= 17,2 mol<br />
n = 112 mol n = 69 mol n = 86 mol<br />
m = n . M<br />
= 112 mol . 32 g/mol<br />
= 3,59 kg
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