Question #6899
A MIXTURE OF POWDERED ALUMINUM AND TIN WAS BURNED IN AN ATMOSPHERE OF OXYGEN IN A WAY SUCH THAT THE RESULTING OXIDES COULD BE COLLECTED AND WEIGHED 0.5488g; THE MIXTURE OF Al2O3 AND SnO2 WEIGHED 0.7712g. CALCULATE THE WEIGHT PERCENT AND ATOM PERCENT OF Al AND Sn IN THE INITIAL MIXTURE
Expert's answer
Weight of Sn is X and weight of Al is (0.5488-x)
x
y
Sn + O2 = SnO2
118.69
150.67
y=x*150.67/118.69=1.2694x
(0.5488-x)
z
4Al + 3 O2 = 2 Al2O3
4*26.98
2*101.93
z=((0.5488-x)*2*101.93)/4*26.98=(0.5488-x)*203.86/107.92=1.0366-1.8889x
y+z=0.7712
1.2694x+1.0366-1.8889x=0.7712
0.6195x=0.2654
x=0.4284
Weight
of Sn is 0.4284g and weight of Al is 0.1204g
x
y
Sn + O2 = SnO2
118.69
150.67
y=x*150.67/118.69=1.2694x
(0.5488-x)
z
4Al + 3 O2 = 2 Al2O3
4*26.98
2*101.93
z=((0.5488-x)*2*101.93)/4*26.98=(0.5488-x)*203.86/107.92=1.0366-1.8889x
y+z=0.7712
1.2694x+1.0366-1.8889x=0.7712
0.6195x=0.2654
x=0.4284
Weight
of Sn is 0.4284g and weight of Al is 0.1204g
