4.83 0.0164 neutralize 0.200 g of NaOH? (c) If 55.8 mL of a BaCl2 solution is ne

Question

4.83

0.0164 neutralize 0.200 g of NaOH? (c) If 55.8 mL of a BaCl2 solution is needed to precipitate all the sulfate ion in a 752-mg sam- ple of Na2S04, what is the molarity of the BaCl2 solution? (d) If 42.7 mL of 0.208 MHCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution? CO, and dditive. ollowing M KMnO 0 M ZnCl2 150.0mL additive. 4.83 Some sulfuric acid is spilled on a lab bench. You can neutral- many mil- 1000.0 mL f the stock what will ize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: 2 NaHCO3(s)+H SO4(ag)Na2SO4(aq)+ 2H20(0)+ 2CO2(8) 0 MHNO M HNO3? l volume of solution? Sodium bicarbonate is added until the fizzing due to the for- mation of CO2(g) stops. If 27 mL of 6.0 MH2SO, was spilled, what is the minimum mass of NaHCO, that must be added to the spill to neutralize the acid? The distinctive odor of vinegar is due to acetic acid, CH,COOH, which reacts with sodium hydroxide according to: CH,COOH(aq) + NaOH(aq)-? incer cells. 9 M, and ontaining hat is the 4.84

Explanation / Answer

moles of H2SO4 = 27 x 6 / 1000 = 0.162

2 NaHCO3 (s) + H2SO4 (aq)   --------------> Na2SO4 (aq) + 2 H2O (l) + 2 CO2 (g)

2 mol NaHCO3 ------------> 1 mol H2SO4

    ??               -------------> 0.162 mol H2SO4

moles of NaHCO3 = 0.162 x 2 / 1 = 0.324 mol

mass of NaHCO3 = 0.324 x 84 = 27.2 g

mass of NaHCO3 = 27.2 g

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