Calculați compoziția procentuală a următoarelor substanțe:

CH4→metan

MgCl→clorură de magneziu

Al2O3→oxid de aluminiu

Trebuie cu rezolvare!!!

MCH4 = AC + 4AH = 12+4×1 = 16g/mol

% C = mC/MCH4×100 = 12/16×100 = Calculeaza !
%H = mH/M×100 = 4/16×100 =

MMgCl2 = AMg + 2ACl = 24+ 2×35,5 = 95g/mol

%Mg = mMg/MMgCl2×100 = 24/95×100 = Calculeaza !
% Cl = mCl/M×100 = 71/95×100 = Calculeaza !

MAl2O3 = 2×AAl + 3×AO = 2×27+3×16 = 102g/mol

% Al = mAl /M×100 = 54/102×100 =

% O = 48/102×100 =

[tex] CH_{4} [/tex]

[tex]M_{ CH_{4} }=12+4\cdot 1=12+4=16[/tex]

[tex]16g CH_{4}....... 12g C.........4gH[/tex]
[tex]100g CH_{4} ..........xgC............ygH[/tex]

[tex]x= \dfrac{100\cdot 12}{16} =75 [/tex]%C

[tex]y= \dfrac{100\cdot 4}{16} =25[/tex]%H


[tex] MgCl_{2} } [/tex]

[tex]M_{MgCl_{2}}=24+2\cdot 35,5=24+71=95[/tex]

[tex]95gMgCl_{2}.......... 24gMg..........71gCl [/tex]
[tex]100gMgCl_{2}............xgMg...........ygCl[/tex]

[tex]x= \dfrac{100\cdot 24}{95} =25,26[/tex]%Mg

[tex]y= \dfrac{100\cdot 71}{95} =74,73[/tex]%Cl

[tex] Al_{2}O_{3} [/tex]

[tex] M_{ Al_{2}O_{3}}=2\cdot27+3\cdot 16=54+48=102[/tex]

[tex]102g Al_{2} O_{3}..........54gAl.........48gO[/tex]
[tex]100g Al_{2} O_{3}...........xgAl...........ygO [/tex]

[tex]x= \dfrac{100\cdot54}{102} =52,94[/tex]%Al

[tex]y= \dfrac{100\cdot48}{102} =47,05[/tex]%O

Sper să-ți fie de folos.