Question

How many milliliters of a 5.20 MHCl

solution are required to react with 45.0 mL
of a 2.05 MAl(OH)3
solution?
3HCl(aq)+Al(OH)3(s)→3H2O(l)+AlCl3(aq)

Answer 1

Step-by-step explanation:

We need (i) a stoichiometrically balanced equation......

Z

n

(

s

)

+

2

H

C

l

(

a

q

)

→

Z

n

C

l

2

(

a

q

)

+

H

2

(

g

)

↑

⏐

And (ii) equivalent quantities of metal and hydrochloric acid.

Moles of zinc

=

8.75

⋅

g

65.39

⋅

g

⋅

m

o

l

−

1

=

0.134

⋅

m

o

l

.

Moles of HCl

=

0.134

⋅

m

o

l

×

2

4.00

⋅

m

o

l

⋅

L

−

1

×

10

3

⋅

m

L

⋅

L

−

1

=

67

⋅

m

L