1)Nitric Acid : 5.2mL Copper:1.046 Cu(s)+4HNO 3 (aq)-->Cu(NO 3 ) 2 +2NO 2 (g)+2H
ID: 680969 • Letter: 1
Question
1)Nitric Acid : 5.2mL Copper:1.046 Cu(s)+4HNO3(aq)-->Cu(NO3)2+2NO2(g)+2H2O(l) Based on the amounts of copper metal and nitricacid used in the first reaction, calculate the number ofHNO3 there are in excess. Concentrated nitric acidhas a concentration of 15.8M. 2) Sodium Carbonate:3.958 g2nd Reaction: 2HNO3(aq)+Na2CO3(s)--->H2O(l)+CO2(g)+2NaNO3(aq) 3rd Reaction :Cu(NO3)2(aq)+Na2CO3(s)-->CuCO3(s)+2NaNO3(aq) Using the moles of HNO3 youcalculated and the moles of Cu(NO3)2 producedfrom the first reaction, calculate the total mass of sodiumcarbonate needed for the second and third reactions. Did you addenough sodium carbonate in the experiment? 1)Nitric Acid : 5.2mL Copper:1.046 Cu(s)+4HNO3(aq)-->Cu(NO3)2+2NO2(g)+2H2O(l) Based on the amounts of copper metal and nitricacid used in the first reaction, calculate the number ofHNO3 there are in excess. Concentrated nitric acidhas a concentration of 15.8M. 2) Sodium Carbonate:3.958 g
2nd Reaction: 2HNO3(aq)+Na2CO3(s)--->H2O(l)+CO2(g)+2NaNO3(aq) 3rd Reaction :Cu(NO3)2(aq)+Na2CO3(s)-->CuCO3(s)+2NaNO3(aq) Using the moles of HNO3 youcalculated and the moles of Cu(NO3)2 producedfrom the first reaction, calculate the total mass of sodiumcarbonate needed for the second and third reactions. Did you addenough sodium carbonate in the experiment?
Explanation / Answer
Moles of nitric acid = molarity * volume = 15.8 M * 0.0052 L =0.08216 mol Moles of copper = 1.046 g / 63.54 g/mol =0.0164 mol From the equation, we know 1 mole of copper reacts with 4moles of nitric acid. 0.0164 moles of copper reacts with 0.0658 moles ofnitric acid. Excess moles of nitric acid = 0.08216 - 0.0658 = 0.0163molesFor the second and third reactions, follow the same procedureusing the moles of nitric acid and copper nitrate. Since moles of nitric acid = 0.0163, moles of sodium carbonate required = 2 * 0.0163 = 0.0326mol Moles of coper nitrate formed = moles of copper reacted =0.0164 mol So moles of sodium carbonate required = 0.0164 Total moles of sodium carbonate = 0.0326+ 0.0164 = 0.049 Mass of sodium carbonate required = 0.049 * 105.99 g/mol =5.19 g The amount of sodium carbonate added is insufficient. For the second and third reactions, follow the same procedureusing the moles of nitric acid and copper nitrate. Since moles of nitric acid = 0.0163, moles of sodium carbonate required = 2 * 0.0163 = 0.0326mol Moles of coper nitrate formed = moles of copper reacted =0.0164 mol So moles of sodium carbonate required = 0.0164 Total moles of sodium carbonate = 0.0326+ 0.0164 = 0.049 Mass of sodium carbonate required = 0.049 * 105.99 g/mol =5.19 g The amount of sodium carbonate added is insufficient.
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