One step in the isolation of pure rhodium metal (Rh) is the precipitation of rho
ID: 500128 • Letter: O
Question
One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation: Rh_2(SO_4)_3(aq) + 6NaO_(aq) 2Rh(OH)_3 (s) + 3Na_2SO_4(aq) If 8.40 g of rhodium(III) sulfate reacts with excess sodium hydroxide, what mass of rhodium(III) hydroxide may be produced? a. 5.23 g b. 16.8 g c.8.4g d. 1.3 g e. 10.4 g Magnesium reacts with iodine gas at high temperatures to form magnesium iodide. What mass of MgI_2 can be produced from the reaction of 5.15 g Mg and 50.0 g I_2? a. 29.5 g b. 44.9 g c. 54.8g d. 55.2 g e.58.9g Aspirin (C_9H_8O_4) is produced by the reaction of salicylic acid (M = 138.1 g/mol) and acetic anhydride (M = 102.1 g/mol). C_7H_6O_3(s) + C_4H_6O_3(l) C_9H_8O_4(s) + C_2H_4O_2(l) If you react 2.00 g C_7H_6O_3 with 1.60 g C_4H_6O_3, what mass of aspirin (M = 180.2 g/mol) can theoretically be obtained? a. 0.40 g b. 2.61 g c. 2.82 g d. 1.53 g e. 3.60 g Aspirin is produced by the reaction of salicylic acid (M = 138.1 g/mol) and acetic anhydride (M = 102.1 g/mol). C_7H_6O_3(s) + C_4H_6O_3(l) C_9H_8O_4(s) + C_2H_4O_2(l) If 2.04 g of C_9H_8O_4 (M = 180.2 g/mol) is produced from the reaction of 3.03 g C_7H_6O_3 and 4.01 g C_4H_6O_3, what is the percent yield? a. 28.8% b. 29.0% c. 50.9% d. 51.6% e.67.3%Explanation / Answer
Molar mass of Rh2(SO4)3 = 103*2 + ( 32 + 16*4 )*3
=494 g/mol
Molar mass of Rh(OH)3 = 103 + ( 16 + 1)*3
= 154 g/mol
Rh2(SO4)3 (aq.) + 6NaOH (aq) ---------> 2Rh(OH)3 (s) + 3Na2SO4 (aq.)
According to stoichiometric coefficient of equation 1 mole or 494g of Rh2(SO4)3 produces 2 mole or 2 *154g of Rh(OH)3.
Mass of rhodium hydroxide produced by 8.40g of rhodium sulfate = (2*154/494)*8.4
= 5.23g
So option (a) is correct.
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