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y-l mol KNO3
132x+101y=100%
39y=K%
28x+14y=N%
K%=N%
39y=28x+14y⇒25y=28x
y=(28x)/25
132x+101y=1
132x+ (101*28*x)/25 =1
132x+113,12x=1
245,12x=1
x=1/245,12
x=25/6128
(132*25)/6128+101y=1
101y=1- 825/1532
101y=707/1532
y=7/1532
n(NH4)2SO4:nKNO3=25/6128 : 7/1532 = 25:28
⇒m(NH4)2SO4:mKNO3=3300:2828=825:707≈132:113
możesz sprawdzić na kalkulatorze i132:113 jest przybliżeniem 825:707