Question

Calculați compoziția procentuală a următoarelor substanțe:

CH4→metan

MgCl→clorură de magneziu

Al2O3→oxid de aluminiu

Trebuie cu rezolvare!!!

Answer 1

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 =

Answer 2

[tex] CH_{4} [/tex]

$M_{ CH_{4} }=12+4\cdot 1=12+4=16$

$16g CH_{4}....... 12g C.........4gH$
$100g CH_{4} ..........xgC............ygH$

$x= \dfrac{100\cdot 12}{16} =75$%C

$y= \dfrac{100\cdot 4}{16} =25$%H


$MgCl_{2} }$

$M_{MgCl_{2}}=24+2\cdot 35,5=24+71=95$

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

$x= \dfrac{100\cdot 24}{95} =25,26$%Mg

$y= \dfrac{100\cdot 71}{95} =74,73$%Cl

$Al_{2}O_{3}$

$M_{ Al_{2}O_{3}}=2\cdot27+3\cdot 16=54+48=102$

$102g Al_{2} O_{3}..........54gAl.........48gO$
$100g Al_{2} O_{3}...........xgAl...........ygO$

$x= \dfrac{100\cdot54}{102} =52,94$%Al

$y= \dfrac{100\cdot48}{102} =47,05$%O

Sper să-ți fie de folos.