Shown below are the 4 reactions involved in the Cu cycle. Cu( s ) + 4 HNO 3 ( aq

ID: 507568 • Letter: S

Question

Shown below are the 4 reactions involved in the Cu cycle.

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

3 Cu(NO3)2(aq) + 2 Na3PO4(aq) Cu3(PO4)2(s) + 6 NaNO3(aq)

Cu3(PO4)2(s) + 6 HCl(aq) 3 CuCl2(aq) + 2 H3PO4(aq)

CuCl2(aq) + Mg(s) MgCl2(aq) + Cu(s)

The first reaction in the cycle is initiated by adding 6 M nitric acid to elemental copper.

1) Calculate the volume of 6 M HNO3 needed to react with 0.636 grams of Cu.

a) Convert grams Cu to moles Cu

0.636gCu X (1molCu/63.546gCu)= 0.010mol Cu

b) Calculate the moles HNO3 needed for the amount of Cu in part (a) using the balanced chemical equation.

0.010mol Cu X (4 mol HNO3/ 1mol Cu)= 0.040mol HNO3

c) Calculate the volume of 6 M HNOneeded to provide the moles of HNOin part (b).

(0.040 mol HNO3/6M)= 0.007 L

d) Calculate yield in movles of the copper(II) nitrate that you expect to obtain from 0.636 grams of copper

0.01 mol Cu(NO3)2

You have previously calculated the amount of 6.00 M nitric acid needed to react with the amount of copper you will be using in this week's lab. You have also calculated the amount of copper(II) nitrate you will make in the first reaction.

It is now time to calculate the amounts of the other reactants needed to complete the copper cycle.

The following reactants are available to carry out the rest of the reactions in the copper cycle.

elemental magnesium

6.00 M hydrochloric acid

0.500 M sodium phosphate

(a) For the second reaction in the cycle, use the moles of copper(II) nitrate expected at the end of the first reaction in the cycle as the starting point for the next calculation.

(i) Calculate the volume in mL of the appropriate reactant needed for the next reaction in the cycle.

_____ mL of 0.500 M sodium phosphate (entered 1.093g but got this wrong)

(ii) Calculate the yield in moles of the copper salt produced by the 2nd reaction.

3.33e-3 mol

(b) For the third reaction in the cycle, use the moles of copper product expected at the end of the second reaction in the cycle as the starting point for the next calculation.

(i) Calculate the volume in mL of the appropriate reactant needed for the third reaction in the cycle.

3.33 mL of 6.00 M hydrochloric acid

(ii) Calculate the yield in moles of the copper salt produced by the 3rd reaction.

9.99e-3 mol

(c) For the final reaction in the cycle, use the moles of copper product expected at the end of the third reaction in the cycle as the starting point for the next calculation.

(i) Calculate the mass in grams of the appropriate reactant needed for the fourth reaction in the cycle.

____ g of elemental magnesium (entered 0.16g and got this wrong)

(ii) Calculate the yield in moles of the solid copper produced by the 4th reaction.

____ mol (entered 3.33e-3mol and got this wrong)

(iii) Calculate the yield in grams of solid copper that should be obtained.

____ g Cu (entered 0.212g and got this wrong)

Then show calculation for the colume in mL of 6.00 M HNO3 needed to react with the amount of copper

Volume = (0.636 g Cu)(1 mol Cu / ___ g Cu)(4 mole HNO3 / 1 mole Cu)(L solution / ____ mol HNO3)(1000ml / 1L)

= _____ mL of 6.00 M HNO3

Explanation / Answer

(a) For the second reaction in the cycle, 3 Cu(NO3)2(aq) + 2 Na3PO4(aq) 1 Cu3(PO4)2(s) + 6 NaNO3(aq)

(i) The moles of copper(II) nitrate expected at the end of the first reaction in the cycle = 0.01 mol

From the balanced chemical reaction, it is clear that, three moles of copper(II) nitrate is required to react with two moles of sodium phosphate. SInce we have 0.01 mol of copper(II) nitrate, therefore

number of mole of sodium phosphate required is = (2 / 3 ) x 0.01 = 0.007 mol

0.500 M = 0.007 mol / Volume of sodium phosphate (L)

Volume of sodium phosphate (L) = 0.007 / 0.500 = 0.013 L

Volume of sodium phosphate (mL) = 0.013 x 1000 = 13.33 mL

(ii) Yield in moles of the copper salt produced by the 2nd reaction:

From the balanced chemical reaction, it is clear that, three moles of copper(II) nitrate form a single mole of copper salt.

3 Cu(NO3)2(aq) + 2 Na3PO4(aq) 1 Cu3(PO4)2(s) + 6 NaNO3(aq)

Therefore, 0.01 mol of copper(II) nitrate will produce = ( 1 / 3 ) x 0.01 mol = 0.0033 mol = 3.33 x 10-3 mol

(b) For the third reaction in the cycle, Cu3(PO4)2(s) + 6 HCl(aq) 3 CuCl2(aq) + 2 H3PO4(aq)

(i) The moles of copper(II) salt expected at the end of the second reaction in the cycle = 3.33 x 10-3mol

From the balanced chemical reaction, it is clear that, six moles of hydrochloric acid is required to react with one moles of copper phosphate. SInce we have 3.33 x 10-3mol of copper salt, therefore

number of mole of hydrochoric acid required is = (6 / 1 ) x 3.33 x 10-3 = 0.02 mol

6.0 M = 0.02 mol / Volume of hydrochloric acid (L)

Volume of hydrochloric acid (L) = 0.02 / 6.0 = 0.003 L

Volume of sodium phosphate (mL) = 0.003 x 1000 = 3.33 mL

(ii) Yield in moles of the copper salt produced by the 3rd reaction:

From the balanced chemical reaction, it is clear that, three moles of copper(II) nitrate form a six moles of hydrochloric acid.

Cu3(PO4)2(s) + 6 HCl(aq) 3 CuCl2(aq) + 2 H3PO4(aq)

Therefore, 0.02 mol of hydrochloric acid will produce = ( 3 / 6 ) x 0.02 mol = 0.01 mol ~ 9.99 x 10-3 mol of copper salt

(i) The moles of copper(II) salt expected at the end of the third reaction in the cycle = 9.99 x 10-3mol

From the balanced chemical reaction, it is clear that, one mole of elemental magnesium is required to react with one mole of copper salt. SInce we have 9.99 x 10-3mol of copper salt, therefore

number of mole of elemental magnesium required is = (1 / 1 ) x 9.99 x 10-3 = 9.99 x 10-3 mol

(ii) Yield in moles of the solid copper produced by the 4th reaction:

From the balanced chemical reaction, it is clear that, one mole of solid copper form from one mole of copper salt.

CuCl2(aq) + Mg(s) MgCl2(aq) + Cu(s)

Therefore, 9.99 x 10-3 mol of copper salt will produce = ( 1 / 1 ) x 9.99 x 10-3 mol = 9.99 x 10-3 mol of solid copper.

(iii) The yield in grams of solid copper:

The atomic weight (i.e. weight of one mole) of solid copper is 63.5 g/mol

therefore, the weight of 9.99 x 10-3 mol of solid copper = 9.99 x 10-3 mol x 63.5 g/mol = 0.635 g

Then, using above calculations in (i), (ii) and (iii), the volume of HNO3 required to react with given 0.0636 g of Cu in equation (i), produces 0.0636 of solid copper after completing the set of four reactions

Therefore,

Volume = (0.636 g Cu)(1 mol Cu / 0.636 g Cu)(4 mole HNO3 / 1 mole Cu)(L solution / 0.040 mol HNO3)(1000ml / 1L)

= 7.000 mL of 6.00 M HNO3

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Shown below are the 4 reactions involved in the Cu cycle. Cu( s ) + 4 HNO 3 ( aq

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