Hi there! I would like some help on the following question. 0.5L of dilute HNO3

Question

Hi there! I would like some help on the following question.

0.5L of dilute HNO3 with unknown concentration is added into 27.2g of a mixture of Cu and Cu2O. The solid material is completely dissolved, forming NO gas and Cu(NO)3. After that, 1.0 L of NaOH solution (1.0M) is added into the above solution.
The Cu2+ ions are completely precipitated forming 39.2 of Cu(OH)2, and the solution is neutral.

(a) Write down all balanced chemical reaction involved in the process.

My answer (not yet balanced) :
HNO3 + Cu + Cu2O -> NO + Cu(NO3)
NO + Cu(NO3) + NaOH -> Cu(OH) + Na + NO4

(b) Calculate the moles of Cu and Cu2O in the initial solid mixture.

(c) Calculate the concentration of HNO3.

Thank you and much appreciated!

2 O.S L of dilute HNOs with unknown concentration is added into 27.2 g of a mixture of Cu and Cu O. The solid material is completely dissolved, forming NO gas and Cu(NOb. After that, 1.0 L of NaOH solution (1.0 M) is added into the above solution. The Cu ions are completely precipitated forming 39.2 g of Cu(OH) and the solution is neutral. (a) Write down all the balanced chemical reactions involved in this process. (4 marks) (4 marks) (4 marks) (b) Calculate the moles of Cu and CusO in the initial solid mixture. () Calculate the concentration of HNO.. Write the Lewis structures and predict using the VSEPR model the geometries of the following species: NH4, NH2 (6 marks) END OF PAPER

Explanation / Answer

a) The balanced equation is

HNO3 + Cu2O = Cu(NO3)2 + H2O + NO

or

NO + 2 Cu2O + 1.4802973661669E-16 HNO3 = 3 Cu + Cu(NO3)

3 NaOH + 3 Cu(NO3) = NO + 3 Cu(OH) + 3 Na + 2 NO4

b) 2H2 + CuO + Cu2O ------] 3Cu +2H2O
use this equation and use the formula moles= mass/ atomic mass of copper solid.

(27.2g/64)/3 and multiply by the atomic mass of CuO because they are in a 3:1 ratio.

Therefore, 0.1416 moles of Cu and Cu2O is in the initial solid mixture.

c) 0.5 litre dilute nitric acid
mol. wt of HNO3 is 72.
take mass percent of HNO3 is 69% so in 500 ml , 69% is HNO3 ie. 345 ml is HNO3.
now , hence no of moles of HNO3 = 345/72 = 4.8
which is nearly 5 moles per litre.

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