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calculate the volume of concentrated nitric acid (16 m) required to react with 0.41 g of cu(s) if you use 1.86 ml (an excess) of hno3, what volume of 5.0m naoh will you need to neutralize the solution after the copper is completely dissolved?

Respuesta :

the volume of concentrated nitric acid (16 m) required to react with 0.41 g of cu(s) if you use 1.86 ml (an excess) of hno3, what volume of 5.0m naoh will you need to neutralize the solution after the copper is completely dissolved is 0.704 ml

4 HNO3(l) + Cu(s) ==> Cu(NO3)2(s and aq) + 2 NO2(g) + 2 H2O(l)

According to net ionic equation 4 moles of HNO3 reacts with 1 mole of Cu(s)

   moles of Cu in 0.36 gm of Cu = 0.36/63.5

moles = molarity*volume(L)

moles of Cu*4 =moles of HNO3

Let volume of HNO3 be x

0.36/63.5*4 = 16*x

x = 1.42 ml

c. Volume of HNO3 added = 1.64 ml

excess of HNO3 = 1.64-1.42

= 0.22 ml

Normality of NaOH and HNO3 is same i.e. equal moles of NaOH and HNO3 neutralises itself

molarity of NaOH = 5 moles/litre

moles of NaOH = Moles Of HNO3

5*x = 16*0.22

x = 0.704 ml

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