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

c) Excess of Na2CO3

We need to calculate how many Na2CO3 the reaction would need to produce 2.65g of CuCl.

2.65g CuCl (1 mol CuCl / 99g CuCl) (1 mol CuCl2/ 2 mol CuCl) (1 mol CuCO3 / 1 mol CuCl2) (1 mol Na2CO3 /1 mol CuCO3) (106g Na2CO3 / 1mol Na2CO3) = 1.419g Na2CO3

2.65g CuCl (1 mol CuCl / 99g CuCl) (1 mol CuCl2/ 2 mol CuCl) (1 mol CuCO3 / 1 mol CuCl2) (2 mol NaNO3 /1 mol CuCO3) (1 mol Na2CO3 / 2mol NaNO3) (106g Na2CO3 / 1mol Na2CO3) = 1.419g Na2CO3

So, the theorical use is 2.838g Na2CO3 and we used 4.40g which represents a 1.562g of excess (55.04% excess).

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