We did the following reactions: 1) Cu(s) + 4HNO3(aq) ? Cu(NO3)2(aq) + 2NO2(g) +

Question

We did the following reactions: 1) Cu(s) + 4HNO3(aq) ? Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) 2) 2HNO3(aq) + Na2CO3(s) ? H2O(l) + CO2(g) + 2NaNO3(aq) This step was done with step three and got rid of the excess HNO3 3) Cu(NO3)2(aq) + Na2CO3(s) ? CuCO3(s) + 2NaNO3(aq) 4) CuCO3(s) + 2HCl ? CuCl2(aq) + H2O(l) + CO2(g) 5) CuCl2(aq) + Cu(s) ? 2CuCl(s) a) How many grams of CuCl(s) can you theoretically produce from 3.50 grams of Cu(s)? What is the % Yield if you only made 8.37 grams? (Cu = 63.6 g/mol and CuCl = 99.0g/mol)

Explanation / Answer

we have number of moles of cu =3.5/63.6=0.055 now by stoichiometry 1mole of cu will prepare 2 moles of cucl so 0.055 moles will prepare 0.11 moles thus number of grams will be =0.11 x 99=10.89 % yield =8.37/10.89 =76.85%

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