Zinc reacts with acids to produce H_2 gas. If you have 52.0 g of Zn, what volume

Question

Zinc reacts with acids to produce H_2 gas. If you have 52.0 g of Zn, what volume of 3.05 M HCl is needed to convert the Zn completely? The reaction of HCl with NaOH is represented by the equation What volume of 0.575 M HCl is required to titrate 42.8 mL of 0.334 M NaOH? Aluminum sulfate (Al_2(SO_4)_3) is mainly used as a flocculating agent in the purification of drinking water and waste water treatment plants. What mass of aluminum sulfate is required to prepare 250.0 mL of 0 627 M Al_2(SO_4)_3 solution?

Explanation / Answer

8)

The reaction which takes place is 2HCl + Zn --> ZnCl2 + H2.
We need to find out first the number of moles of Zn reacting with HCl.
To do this we take the molar mass of Zn which is 65.38 g/mol. We have 52 g of Zn, so divide 52g by 65.38 g/mol to get 0.795 moles of Zn in the reaction.
Now we need to use this mole amount to find the mole amount of HCl needed. Because this reaction requires twice as many moles of HCl as it does Zn, we will need 0.795 x 2 or 1.59 moles of HCl.
Now we can use this mole amount and the molarity to find the volume of HCl (aq) needed.
Remember, molarity (M) = moles/L, therefore L = moles/molarity. We need 1.59 moles so divide that by 3.05 M and you get 0.521 L of 3.05 M HCl needed. 1000mL = 1 L so 0.521 L = 521 mL.

Summary:
521 ml of 3.05 M HCl is needed to react with 52 g of Zn.

9)

We Know that N1V1 = N2V2

N1 = molarity of HCl = 0.575 M

N2 = molarity of NaOh = 0.334 M

V1= Volme of HCl =  ?

V2 = Volme of NaOH = 42.8 ml

So putting these values in the equation

we get

0.575 * V1 = 0.334 * 42.8

=> V1 = 0.334 * 42.8 / 0.575 = 24.86 ml of HCl

10)

Molarity = moles / lit

so 0.627 M = 0.627 moles / lit

= 0.627 / 4 / lit/4 = 0.156 moles / 250 ml

So preparation of 0.627 M of 250 ml Al2(SO4)3 we required 0.156 moles of aluminium sulfate

= 0.156 * 342.15 = 53.37 gm of aluminium sulfate

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