Part IX Part XI What is the percentage ved of o,#2as g of Koo. (molar mass 123 g

Question

Part IX Part XI What is the percentage ved of o,#2as g of Koo. (molar mass 123 g) is decomposed to produced 3.2 gof O, (molar mass 32 g) according to the equation above? A) 100% b) 67% c)50% d) 33% Permanganate and oxalate ions react in an acidified solution according to the balanced equation above. How many moles Moriei are produced when 20. ml of acidified 0.20 M KMnO, solution are added to s0. ml of 0.10 M Na.C.O? )0 0020 mol b)00040 mol )0.0090 mo d) 0.010 mol Part X in the reaction represented above, what mass of Hi is produced by the reaction of 2.0 x 10 molecules H, with excess 1,? a)94g b) 85g c)64g d)43 g 67. What is the empirical formula of an oxide of sulfur f the compound contains 16 g of sulfur for every 24 g of oxygen? 68. A student prepares a solution by dissolving 90.0g glucose (molar mass 180.2 s mol-J in enough water to should be reported as )2.00 M b)2 000 M c)8.00 M d) 8.000 M 76.f 100. mL of 1.0 M NaOH is diluted with distilled water to a volume of 2.0L, the concentration of the resulting solution is a) 0.025 M b)0.050 M c)0.10 M d) 20.M

Explanation / Answer

63.

2 KClO3 (s) ------------> 2 KCl (s) + 3 O2 (g)

Mass of KClO3 = 24.6 g.

Molar mass of KClO3 = 123 g/mol

Moles of KClO3 = mass / molar mass = 24.6 / 123 = 0.200 mol

From the balanced equatio,

2 mol of KClO3 produces 3 mol of O2

Then,

0.200 mol of KClO3 produces 0.300 mol of O2

So,

Theretical Mass of O2 produced = moles * molar mass = 0.300 * 32.0 = 9.60 g.

% Yield = Actual yield * 100 / Theroretical yield = 3.20 * 100 / 9.60 = 33.3

(d) 33 % is the answer

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