Question

9 of 14 Constants | Periodic Table You may want to reference (Pages 295 - 299)...

9 of 14

Constants | Periodic Table

You may want to reference (Pages 295 - 299) Section 7.4 while completing this problem.

Sulfuric acid dissolves aluminum metal according to the following reaction:

2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)

Suppose you wanted to dissolve an aluminum block with a mass of 15.7 g .

Part A

What minimum mass of H2SO4 would you need?

m =

81.1481.14

g

Incorrect; Try Again; 5 attempts remaining

Part B

What mass of H2 gas would be produced by the complete reaction of the aluminum block?

m =

nothing

g

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A)

Molar mass of Al = 26.98 g/mol

mass of Al = 15.7 g
mol of Al = (mass)/(molar mass)
= 15.7/26.98
= 0.5819 mol

According to balanced equation
mol of H2SO4 reacted = (3/2)* moles of Al
= (3/2)*0.5819
= 0.8729 mol

Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol

mass of H2SO4 = number of mol * molar mass
= 0.8729*98.09
= 85.62 g

B)

According to balanced equation
mol of H2 formed = (3/2)* moles of Al
= (3/2)*0.5819
= 0.8729 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = number of mol * molar mass
= 0.8729*2.016
= 1.76 g

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