Balance the following reaction. What amount of H_2SO_4 is necessary to completel

Question

Balance the following reaction. What amount of H_2SO_4 is necessary to completely dissolve 15.2 g of aluminum? Al() + H_2SO_4 (aq) rightarrow Al_2 (SO_4)_3 (aq) + H_2 (g) Potassium superoxide, KO_2, is used in rebreathing gas masks to generate oxygen. The potassium superoxide reacts with water to produce molecular oxygen and potassium hydroxide (KOH). Write a balanced reaction. If the reaction vessel contains 0.25 mol KO_2 and 0.15 mol water how many grams of oxygen can be produced? Pb^2+ (aq) + 2KCl_(aq) rightarrow PbCl_2 (s) + 2K^+ (aq) Lead ions can be precipitated from solution with KC1 according to the following reaction: When 28.5 g KC1 is added to a solution containing 25.7 g Pb^2+, a PbCl_2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4 g. Determine the limiting reactant, theoretical yield of PbCl_2, and the percent yield of the reaction.

Explanation / Answer

2Al + 3H2SO4 ---------->Al2(SO4)3 + 3H2 ( balanced reaction)

the reaction tells us 2 moles of Al requires 3 mole of H2SO4

atomic weight of Al =27, molar mass of H2SO4= 98

moles of Al in 15.2 gm= 15.2/27= 0.56

hence moles of H2SO4 required = 3*0.56/2= 0.84, mass of H2SO4= moles of H2SO4* molar mass= 0.84*98= 82.32 gm

2. KO2+H2O ------->KOH + O2 ( balanced reaction)

according to the reaction ,molar ratio of KO2: H2O= 1:1

actual ratio of KO2: H2O= 0.25:0.15= 1.66 :1

KO2 is excess and H2O is limting. Moels of O2 produced= 0.15 , mass of O2= moles of O2* molar mass= 0.15*32=4.8 gm

3. The reaction is Pb+2+ 2KCl ------>PbCl2+2K+

atomic weights: Pb = 207, molar mass of KCl= 39+35.5= 74.5

moles= mass/moalr mass (or atomic mass)

molar ratio of Pb+2 : KCl= 1:2

moles of Pb= 25.7/207=0.1241, KCl= 28.5/74.5= 0.38

actual ratio =0.1241:0.38= 1: 3.1

So KCl is excess and all the Pb+2 reacts because it is the liminng reactant. Moles of PbCl2 formed = 0.1241

molar mass of PbCl2= 278, mass of PbCl2= moles* molar mass= 278*0.1241= 34.5 gm

This is the theoreical yield. Actual yield= 29.4 gm

percent yield= 100* actual yield/theoretical yield= 100*29.4/34.5= 85.22%

1 mole of KO2 produces 1 mole of O2. molar masses : KO2= 71 and O2= 32

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