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

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