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

Question

For the following reaction, 3.82 grams of copper are mixed with excess silver nitrate. The reaction yields 5.93 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 process yield of copper(II) nitrate? (ANSWER HERE) For the following reaction 4.59 grams of nitrogen monoxide are mixed with excess hydrogen gas. The reaction yields 1.88 grams of nitrogen gas. Nitrogen monoxide(g) + hydrogen(g) nitrogen(g) + water (l) What is the theoretical yield of nitrogen gas? (ANSWER HERE) grams What is the present yield of nitrogen gas? (ANSWER HERE) % How many grams of PbBr_3 will precipitate when excess ZnBr_3 solution is added to 51.0 mL of 9.703 M Pb(NO_3)_2 solution? Pb(NO_3)_2 (aq) + ZnBr_3 (aq) PbBr_3 (s) + Zn(NO_3)_2 (aq) (ANSWER IN GRAMS) How many mL of 4.557 M HI are needed to dissolve 7.34 g of MgCO_3 2HI(aq) + MgCO_3 (aq) MgI_3 (aq) + H_2 O (l) + CO_2 (g) (ANSWER IN mL) Calculate the number of milliliters of 0.799 M NaOH required to precipitate all of the Ca^3+ ions in 127 mol. of 0.704 M CaSO_4 solutions at Ca(OH)_3. The equation for the reaction is: CaSO_4 (aq) + 2NaOH(aq) Ca(OH)_3 (s) + Na_2 SO_4 (aq) ANSWER IN mL NaOH An aqueous solution of hydroxide acid is standardized by nitration with a 0.156 M solution of sodium hydroxide. If 12.5 mL of base are required to neutralize 28.0 mL of the acid, what is the molarity of the hydroxide acid solution? (ANSWER IN) M hydroxide acid An aqueous solution of barium hydroxide is standardized by nitration with a 1.34 M solution of nitric acid. If 15.9 mL of base are required to neutralize 19.4 mL of the acid, what is the molarity of the barium hydroxide solution? (ANSWER IN) M barium hydroxide An aqueous solution of calcium hydroxide is standardized by nitration with a 0.200 M solution ofhydrochloric acid. If 35.7 mL of base are required to neutralize 25.2 mL of the solid, what is the molarity of the calcium hydroxide solution? (ANSWER IN) M calcium hydroxide

Explanation / Answer

(1) moles of Cu = 3.82/63.546 = 0.061 mols

Theoretical yield of Cu(NO3)2 = 0.061 x 187.56 = 11.275 g

Percent yield = 8.93 x 100/11.275 = 79.20%

(2) moles of NO = 4.59/30.01 = 0.153 mol

Theoretical yield of N = 0.153 x 28 = 4.29 g

Percent yield = 1.88 x 100/4.28 = 43.90%

(3) moles of Pb(NO3)2 = 0.703 M x 51 ml = 35.853 mmol

grams of PbBr2 formed = 35.853 x 367.01/1000 = 13.16 g

(4) moles of MgCO3 = 7.34/84.314 = 0.087 mol

Volume of HI required = 0.087 x 2 x 1000/0.537 = 324.30 ml

(5) moles of CuSO4 = 0.704 M x 127 ml = 89.408 mmol

volume of NaOH required = 89.408 x 2/0.799 = 223.80 ml

(6) moles of NaOH = 0.156 M x 12.5 ml = 1.95 mmol

molarity of HI = 1.95/28 = 0.07 M

(7) moles of HNO3 = 0.134 M x 19.4 ml = 2.6 mmol

molarity of Ba(OH)2 = 2.6/2 x 15.9 = 0.082 M

(8) moles of acid = 0.2 M x 25.2 ml = 5.04 mmol

molarity of Ca(OH)2 = 5.04/2 x 25.7 = 0.098 M

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