Question

sa se calculeze compozitia procentuala a urmatoarelor substante :

-suflat de bariu

-clorula de sodiu

-oxid de aluminiu

-azotat de calciu

-fosfat de calciu

-fosfat de magneziu
sa se procedeze prin cele doua etape

va rog e urgentt

Answer 1

BaSO4

M=137 +32 + 64=233-----> 233g/moli

233g BaSO4-------137 g Ba-------32 g S --------64 g O

100 g-------------------x------------------y------------------z

x=100.137 : 233=58,80% Ba

y=100.32 : 233=13,73 % S

z=100.64 : 233=27,46 % O

NaCl

M=23 +35,5=58,5------>58,5g/moli

58,5g NaCl---------23 g Na---------35,5 gCl

100g------------------xg Na--------------yg Cl

x=100.23 :58,5=39.31 % Na

y=100.35,5 : 58,5=60,68 % Cl

Al2O3

M=2.27 + 3.16=102-----> 102g/moli

102g Al2 O3--------54 g Al--------48 g O

100g---------------------x-------------------y

x=100.54 : 102=52,94 % Al

y=100.48 : 102=47,06 % O

Ca(NO3)2

M=40 + 2.14 +6.16=164------>164g/moli

164 g Ca(NO3)2----------40g Ca ------28g N--------96gO

100g----------------------------x-----------------y-------------------z

x=100.40 :164=24,39% Ca

y=100.28 : 164=17,07% N

z=100.96 :164=58,53 % O

Ca3(PO4)2

M=3.40 +2.31 +8.16=310 ------> 310g/moli

310g Ca3(PO4)2-------120 g Ca ------62 g P---------128 g O

100g--------------------------x----------------y-------------------z

x=100.120 : 310=38,7% Ca

y=100.62 :310=20% P

z=100.128 :310=41,3 % O

Mg3(PO4)2

M=3.24 + 2.31 +8.16=260------>260g/moli

260g Mg3(PO4)2---------72g Mg--------62g P---------128 g

100 g----------------------------x------------------y--------------z

x=100.72 :260=

y=100.62:260=

z=100.128 : 260=

Comp. proc. prin raport de masa

BaSO4

r.m.= 137 : 32 : 64

137+32+64=233

se face la fel cu prima metoda

NaCl

r.m.=23 :35,5

23+35,5=58,5

se face la fel cu prima metoda

Al2o3

r.m.=54: 48=9: 8  [s-a simplif. cu 8]

9+ 8 =17

17 g Al2 O3-------9gAl----------8 g O

100 g------------------x-----------------y

x=100 .9 : 17=52,94 % Al

y=100.8 : 17=47,06 % O

[pe celelalte  te rog sa le lucrezi singur , dupa modele]