1. Al(s) + O2(g) Al2O3(s) (a) Consider the unbalanced equation above. How many g

Question

1. Al(s) + O2(g) Al2O3(s)

(a) Consider the unbalanced equation above. How many grams of O2 are required to react with 32.0 g of aluminum? Use at least as many significant figures in your molar masses as in the data given (_______ g O2)

(b) What mass of Al2O3 is produced? (________g Al2O3 )

2.

(a) Consider the unbalanced equation above. How many grams of O2 are required to react with 47.5 g of C2H6? Use at least as many significant figures in your molar masses as in the data given.
_________g O2
(b) What mass of CO2 is produced?
_________g CO2
(c) What mass of H2O is produced?
_________g H2O

Explanation / Answer

1) a)Balanced eq is 4 Al (s) + 3O2 (g) ---> 2Al2O3 (s)

Al moles = mass of Al/atomic mass of AL = 32 /26.98154 =1.186

O2moles need = ( 3/4) Al moles = ( 3/4) x 1.186 = 0.8895

O2 mass = moles of O2 x molar mass of O2 = 0.8895 x 32 g/mol = 28.46 g

b) Al2O3 moles produced = (2/4) Al moles = ( 2/4) x 1.186 = 0.593

Al2O3 mass = moles x molar mass of Al2O3 = 0.593 x 101.96 = 60.46 g

2) Balanced eq is C2H6 + 3.5 O2 ---> 2CO2 + 3H2O

a) C2H6 moles = 47.5 /30.07 = 1.58

O2 moles = 3.5 x C2H6 moles = 1.58 x 3.5 = 5.53

O2 mass = moles of O2 x molar mass of O2 = 5.53 x 32 = 176.96 g

b) CO2 moles = 2 x C2H6 moles = 1.58 x 2 = 3.16

CO2 mass = moles x molar mass of CO2 = 3.16 x 44.01 = 139.07 g

c) H2O moles = 3 x C2H6 moles = 1.58 x 3 = 4.74

H2O mass = 4.74 x 18 = 85.32 g

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