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LUnt (IMrate ion), and then reduce the copper(II) ion to copper(I) by the te copper(II), remove the addition of excess copper metal. ono to n ovlo The chemical reactions you will carry out in this lab are shown below. 1) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(1) 2il ed 2) 2HNO3(aq) + Na2CO3(s)→ H2O(1) ±ÇO2(g) + 2NaNO3(aq) 0. bay 3) Cu(NO3)2(aq) + NazCO3(s) → ÇUCO:(s) + 2NANO3(aq) 4) CUCO3() + 2HC((aq) CuCh9) + H2O@ + CO2«g) 12 5) CuC2(aq) + Cu(s) 2CUCI) BEPpe to anstmoo adi The reactions will be carried out in such a way that the copper metal added in the first reaction will determine the amount of copper(I) chloride that forms in the last reaction because it is the limiting reagent in all reactions where it's involved. In other words, all the other reagents used for this preparation will be added in excess. The theoretical yield is then based on the amount of copper added in the first reaction. 20 Calculations: cwoda 1n bluoda Theoretical yield = -g CuC) " mole CuCl 1 mole Cu 1 mole Cu(NO3)2 X. 1 mole CuCO3 1 mole CuCl2 2 mole CuCl -g Cu x - - g Cu 1 mole Cu 1 mole Cu(NO3)2 1 mole CuCO3 * 1 mole CuCl2 oLas 1st You need to fill in the molar masses of Cu and of CuCl. Experimental yield x 100 bouos a ni Tuqgo2 ucvoaroo bor 0%Yiel

LUnt (IMrate ion), and then reduce the copper(II) ion to copper(I) by the te copper(II), remove the addition of excess copper metal. ono to n ovlo The chemical reactions you will carry out in this lab are shown below. 1) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(1) 2il ed 2) 2HNO3(aq) + Na2CO3(s)→ H2O(1) ±ÇO2(g) + 2NaNO3(aq) 0. bay 3) Cu(NO3)2(aq) + NazCO3(s) → ÇUCO:(s) + 2NANO3(aq) 4) CUCO3() + 2HC((aq) CuCh9) + H2O@ + CO2«g) 12 5) CuC2(aq) + Cu(s) 2CUCI) BEPpe to anstmoo adi The reactions will be carried out in such a way that the copper metal added in the first reaction will determine the amount of copper(I) chloride that forms in the last reaction because it is the limiting reagent in all reactions where it's involved. In other words, all the other reagents used for this preparation will be added in excess. The theoretical yield is then based on the amount of copper added in the first reaction. 20 Calculations: cwoda 1n bluoda Theoretical yield = -g CuC) " mole CuCl 1 mole Cu 1 mole Cu(NO3)2 X. 1 mole CuCO3 1 mole CuCl2 2 mole CuCl -g Cu x - - g Cu 1 mole Cu 1 mole Cu(NO3)2 1 mole CuCO3 * 1 mole CuCl2 oLas 1st You need to fill in the molar masses of Cu and of CuCl. Experimental yield x 100 bouos a ni Tuqgo2 ucvoaroo bor 0%Yiel

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter21: The Chemistry Of The Main Group Elements
Section: Chapter Questions
Problem 131SCQ: Xenon trioxide, XeO3, reacts with aqueous base to form the xenate anion, HXeO4. This ion reacts...
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Calculate the theoretical yield. My cu is 1.006 g 

LUnt (IMrate ion), and then reduce the copper(II) ion to copper(I) by the te copper(II), remove the addition of excess copper metal. ono to n ovlo The chemical reactions you will carry out in this lab are shown below. 1) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(1) 2il ed 2) 2HNO3(aq) + Na2CO3(s)→ H2O(1) ±ÇO2(g) + 2NaNO3(aq) 0. bay 3) Cu(NO3)2(aq) + NazCO3(s) → ÇUCO:(s) + 2NANO3(aq) 4) CUCO3() + 2HC((aq) CuCh9) + H2O@ + CO2«g) 12 5) CuC2(aq) + Cu(s) 2CUCI) BEPpe to anstmoo adi The reactions will be carried out in such a way that the copper metal added in the first reaction will determine the amount of copper(I) chloride that forms in the last reaction because it is the limiting reagent in all reactions where it's involved. In other words, all the other reagents used for this preparation will be added in excess. The theoretical yield is then based on the amount of copper added in the first reaction. 20 Calculations: cwoda 1n bluoda Theoretical yield = -g CuC) " mole CuCl 1 mole Cu 1 mole Cu(NO3)2 X. 1 mole CuCO3 1 mole CuCl2 2 mole CuCl -g Cu x - - g Cu 1 mole Cu 1 mole Cu(NO3)2 1 mole CuCO3 * 1 mole CuCl2 oLas 1st You need to fill in the molar masses of Cu and of CuCl. Experimental yield x 100 bouos a ni Tuqgo2 ucvoaroo bor 0%Yiel
Transcribed Image Text:LUnt (IMrate ion), and then reduce the copper(II) ion to copper(I) by the te copper(II), remove the addition of excess copper metal. ono to n ovlo The chemical reactions you will carry out in this lab are shown below. 1) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(1) 2il ed 2) 2HNO3(aq) + Na2CO3(s)→ H2O(1) ±ÇO2(g) + 2NaNO3(aq) 0. bay 3) Cu(NO3)2(aq) + NazCO3(s) → ÇUCO:(s) + 2NANO3(aq) 4) CUCO3() + 2HC((aq) CuCh9) + H2O@ + CO2«g) 12 5) CuC2(aq) + Cu(s) 2CUCI) BEPpe to anstmoo adi The reactions will be carried out in such a way that the copper metal added in the first reaction will determine the amount of copper(I) chloride that forms in the last reaction because it is the limiting reagent in all reactions where it's involved. In other words, all the other reagents used for this preparation will be added in excess. The theoretical yield is then based on the amount of copper added in the first reaction. 20 Calculations: cwoda 1n bluoda Theoretical yield = -g CuC) " mole CuCl 1 mole Cu 1 mole Cu(NO3)2 X. 1 mole CuCO3 1 mole CuCl2 2 mole CuCl -g Cu x - - g Cu 1 mole Cu 1 mole Cu(NO3)2 1 mole CuCO3 * 1 mole CuCl2 oLas 1st You need to fill in the molar masses of Cu and of CuCl. Experimental yield x 100 bouos a ni Tuqgo2 ucvoaroo bor 0%Yiel
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