For the following reaction, 3.82 grams of copper arc mixed with excess silver ni

Question

For the following reaction, 3.82 grams of copper arc mixed with excess silver nitrate. The reaction yields 8.93 grams of copper(II) nitrate. silver nitrate (aq) + copper (s) copper(II) nitrate (aq) + silver (s) What is the theoretical yield of copper(II) nitrate ? (ANSWER HERE)grams What is the percent yield of copper(II) nitrate?(ANSWER HERE) % For the following reaction. 4.59 grams of nitrogen monoxide arc mixed with excess hydrogen gas. The reaction yields 1.88 grams of nitrogen gas. nitrogen mohoxide (g) + hydrogen (g) nitrogen (g) + water (l) What is the theoretical yield of nitrogen gas? (ANSWER HERE) grams What is the percent yield of nitrogen gas?(ANSWER HERE) % How many grams of PbBr_2 will precipitate when excess ZnBr_2 solution is added to 51.0 mL of 0.703M Pb(NO_3)_2 solution? Pb(NO_3)_2 (aq) + Zn Br_2 (aq) PbBr_2 (s) + Zn(NO_3)_2 (aq) (ANSWER IN GRAMS) How many mL of 0.537 M HI are needed to dissolve 7.36 g of MgCO_3? 2HI (aq) + MgCO_3 (s) MgI_2 (aq) + H_2 O (I) + CO_2 (g) (ANSWER IN mL) Calculate the number of milliliters of 0.799 M NaOH required to precipitate all of the Co^2+ ions in 127ml. of 0.704 M CoSO_4 solution as Co(OH)_2. The equation for the reaction is: CoSO_4 (aq) + 2NaOH (aq) Co(OH)_2 (s) + Na_2 SO_4 (aq) ANSER IN mL NaOH

Explanation / Answer

1)
Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag

moles of Cu = mass of Cu / molar mass of Cu
= 3.82 g / 63.5 g/mol
= 0.06 mol

moles of Cu(NO3)2 formed = moles of Cu reacted
= 0.06 mol

molar mass of Cu(NO3)2 = 187.6 g/mol

Theoretical yield = moles * molar mass
= 0.06 * 187.6
= 11.29 g

percent yield = 8.93*100/11.29 = 79.12 %

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