Using the following reaction Mg(NO 3 ) 2(aq) + 2 NaOH (aq) --> Mg(OH) 2 (s) + 2

Question

Using the following reaction

Mg(NO3)2(aq) + 2 NaOH (aq) --> Mg(OH)2 (s) + 2 NaNO3 (aq)

To 2.0 mL of 0.10 M Mg(NO3)2 in a test tube, 3.0 M NaOH is added dropwise

a) calculate the number of moles of Mg(NO3)2 in the solution

b) calculate othe volume in mL of NaOH which is needed to reat completley with the Mg(NO3)2

c) if one drop of the NaOH solution ahs a volume of 0.04 mL, how many drops of 3M NaOH are needded to react compeltely with the Mg(NO3)2?

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Explanation / Answer

Answer – We are given [Mg(NO3)2] = 0.10 M , volume = 2.0 mL , [NaOH] = 3.0 M

reaction –

Mg(NO3)2(aq) + 2 NaOH (aq) ----> Mg(OH)2 (s) + 2 NaNO3 (aq)

a) We know the molarity means moles per liter,

so, molarity = moles / volume(L)

moles = molarity * volume (L)

moles of Mg(NO3)2 = 0.10 M * 0.002 L

                                 = 0.0002 moles

So, the number of moles of Mg(NO3)2 in the solution is 0.0002 moles.

b) We need to calculate the moles NaOH first, since we are given molarity of NaOH

from the balanced reaction –

1 moles of Mg(NO3)2 = 2 moles of NaOH

So, 0.0002 moles of Mg(NO3)2 = ?

= 0.0004 moles of NaOH

Now we know,

molarity = moles / volume(L)

so, volume (L) = moles / molarity

                         = 0.0004 moles / 3.0 M

                         = 0.000133 L

We know,

1 L = 1000 mL

So, 0.0000133 L = ?

= 0.133 mL

So, 0.133 mL of NaOH which is needed to react completely with the Mg(NO3)2.

c) We are given, 1 drop = 0.04 mL and we calculated the volume of NaOH which is needed to react completely with the Mg(NO3)2.

So,

0.04 mL = 1 drop

0.133 mL = ?

= 3.3 drops.

So, 3.3 drops of 3M NaOH are needed to react completely with the Mg(NO3)2.

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