The following data were collected for the preparation of the Copper (i) chloride

Question

The following data were collected for the preparation of the Copper (i) chloride in the lab. (4 + 4 + 6) a. Fill in the missing data and show your work clearly. Based on the above data, calculate the number of moles of HNO_3 are in excess. c. Also based on the above data, the excess moles of HNO_3 used and the moles of Cu(NO_3)_2 produced in the first step, show clearly using calculations that the sodium carbonate was added in excess. The following reactions should be useful: i. Cu (s) + 4HNO_3((aq) rightarrow Cu(NO_3)_2 + 2NO_2(g) + H_2O(I) ii. 2NaNO_3(aq) + Na_2CO_3(s) rightarrow H_2O(I) + CO_2(g) + 2NaNO_3(aq) iii. Cu(NO_3)_2(aq) + Na_2CO_3(s) rightarrow CuCO_3(s) + 2NaNO_3(aq) iv. CuCO_3(s) + 2HCl(aq) rightarrow CuCl_2(aq) + H_2O(l) + CO_2(g) v. CuCl_2(aq) + Cu(s) rightarrow 2CuCl(s)

Explanation / Answer

b) Excess of HNO3

For this, we must calculate how much HNO3 is required theorically to produce 2.65g of CuCl (careful to use the experimental yield).

2.65g CuCl (1 mole CuCl/99g CuCl) (1 mol CuCl2 / 2 mol CuCl) (1 mol CuCO3 / 1 mol CuCl2) (2 mol NaNO3 /1 mol CuCO3) (2 mol HNO3 / 2 mol NaNO3) (63g HNO3/ 1 mol HNO3) = 1.686g HNO3

2.65g CuCl (1 mole CuCl/99g CuCl) (1 mol CuCl2 / 2 mol CuCl) (1 mol CuCO3 / 1 mol CuCl2) (1 mol Cu(NO3)2 / 1 mol CuCO3) (4 mol HNO3 / 1 mol Cu(NO3)2 ) (63g HNO3/ 1 mol HNO3) = 3.373g HNO3

So, the theorical usage of HNO3is 5.059g HNO3

Now, we have to calculate how many grams of HNO3 we used, let's convert those 5.5mL of HNO3 15.8M to grams:

5.5 mL HNO3 (15.8 mol HNO3 / 1000mL HNO3) (63g HNO3 / 1 mol HNO3) = 5.475g HNO3

To calculate the excess just substract the HNO3 used from the theorical HNO3

Excess = 5.475g -5.059g

Excess = 0.416g

In %:

% excess = (0.416g/5.059g)*100

% excess = 8.22%

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