Question

Use seratch paper if necessary, and copy your completed calculations to this sheet Shown below are the 4 reactions involved in the Cu cycle listed in random order CuPO )HC1 ag)--CuCl (ag)+HPO, (aq) CuCl (ag)+Mg )-MgCl (ag)+Cu (G Cu )+4HNO, (ag)-CuNO) (ag)+2NO2 ()+2HO D CuNO (ag)+NaPO. (ag)- Cu(PO.)2 (s) +NaNO, (ag)

Answer 1

Given : the first reaction is addition of 6M nitric acid to elemental copper.

That is of the given reaction :

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

Number of moles = molecular mass

1mol of Cu= 63.55 g

1mol of HNO₃ =63.01 g

4 mol of HNO₃ =4*63.01 g = 252.04 g

1mol of Cu(NO₃)₂=187.56 g

1mol of NO₂= 46 g

2mol of NO₂= 2*46 g= 92g

1mol of H₂O= 18 g

2 mol of H₂O= 2*18 g = 36g

1) 6.67 ml of 6 molar nitric acid is needed to react with 0.636 gram copper

In the given reaction:

1 mol of Cu reacts with 4 mol of HNO₃

In terms of molar masses

63.55 g of Cu reacts with 252.04 g of HNO₃

1g of Cu reacts with 252.04/63.55 g of HNO₃

1g of Cu reacts with 3.966g of HNO₃

0.636 g of Cu reacts with 3.966g*0.636  of HNO₃

0.636 g of Cu reacts with 2.522g of HNO₃

mass of HNO₃ to be reacted = 2.522g

Formula to be used

Molarity (M)= number of moles of solute(n)/ Volume of solution in L

M= n/V

given : Molarity of HNO₃ (M)= 6M = 6mol/L

mass of HNO₃ (w)= 2.522 g

Molar mass of HNO₃ (M.M)= 63.01 g g/mol

number of moles(n)= given mass(w)/molar mass(M.M)

number of moles(n) =2.522 g/(63.01 g/mol)

Cancelling unit of Gram from both the numerator and denominator.

number of moles(n) =(2.522 /63.01 )mol

n= 0.040 mol

Putting all values in the given formula

M= n/V

V = n/M

V= 0.04 mol/(6 mol/L)

Cancelling unit of 'mol' from both the numerator and denominator

V= (0.04/6)L

V= 0.006671 L

1L = 1000ml

0.00667 L =( 1000*0.00667)ml = 6.67 ml

V= 6.67 ml

a.0.636 g of Cu =0.010 mol of Cu

given mass of Copper = 0.636 g

Number of moles = molecular mass

1mol of Cu= 63.55 g

63.55g Cu = 1mol of Cu

1g = 1/63.55 mol of Cu

(Given mass = 0.636 g )

0.636 g = 0.636g*(1/63.55g) mol of Cu

Cancelling unit of Gram from both the numerator and denominator.

0.636 g =(0.636/63.55) mol of Cu

0.636 g =0.010 mol of Cu

b. 0.040 mol of HNO₃ are needed for the amount of copper in part (a) using the balance chemical equation.

According to balance chemical reaction 1 mole of copper reacts with 4 mol of nitric acid.

In part (a) number of moles of copper are calculated equals to 0.010 mole of copper.

1 mole of Cu reacts with 4 mol of HNO₃

0.010 mol of Cu reacts with 0.010*4 mol of HNO₃

0.010 mol of Cu reacts with 0.040 mol of HNO₃

c. 6.67 ml of 6 M HNO₃ need to provide the moles of HNO3 in part-b

Moles of HNO₃ from part (b) (n)= 0.040 mol

Molarity of HNO₃ (M)=6M

M= n/V

V = n/M

V = 0.040 mol /6M

( Units 1M = 1mol/L)

V = 0.040 mol /6mol/L

Cancelling unit of 'mol' from both the numerator and denominator

V= (0.040/6) L

V= 0.006666 L

V= 0.667 L

1L = 1000ml

0.00667 L =( 1000*0.00667)ml = 6.67 ml

V= 6.67 ml

d. 0.0157 mol Copper(||) nitrate expected to obtain from 0.636 gram copper

From the balanced chemical reaction

1 mol Cu gives 1mol Copper(||) nitrate

In terms of mass

63.55g Cu gives 187.56g Copper(||) nitrate

1g Cu gives 187.56/63.55g Copper(||) nitrate

1g Cu gives 2.95 g Copper(||) nitrate

Moles of Copper(||) nitrate = mass /molar mass

Moles of Copper(||) nitrate = 2.95 /187.56= 0.0157 mol