answer both 1. A 15.6g sample of sodium bicarbonate (NaHCO,) is decomposed by be

Question

answer both

1. A 15.6g sample of sodium bicarbonate (NaHCO,) is decomposed by beating How many grams ofNaCO, remains? 2NaHC)(s) NayCO, (s) + CO2 (g) + H2O (g) How many combined grams of CO, and H:0 in qaestion 1? a. 2. A 4.832g mixture of KCIO, and NaCl is heated on a hot plate causing the decomposition to form KC1 (s) and Op (g) IE3474 g of the mixture remains afier complee decomposition: 2KOOs (s) 2KCI (s) + 3O2(g) a. Calculate moles of O2 lost. b. Calculate the grams of KCIO, present in the initiail mixture. c. Caleulate the percent composition of KCO0, in the mixture

Explanation / Answer

a) Mass of NaHCO3= 15.6 gms Molecular weight of NaHCO3= 23+1+12+48= 84

Moles of NaHCO3= mass/molecular weight = 15.6/84=0.186 moles

From the reaction 2NaHCO3---> Na2CO3+ H2O+CO2

2 moles of NaHCO3 gives 1 mole of Na2CO3, 1 mole of CO2 and 1 mole of H2O

0.185 moles give 0.186/2= 0.093 moles of Na2CO3 and 0.093 mole of H2O and 0.093 mole of CO2

molecular weight of Na2CO3= 23*2+12+48= 106

Mass of Na2CO3= 106*0.093=9.858 gm of Na2CO3.

Mass of H2O =0.093*18=1.674 gm and Mass of CO2= 0.093*44=4.092 gms

total mass of CO2 and H2O= 1.674+4.092=5.766 gms

2.

Since only KClO3 undergoes decomposition, hence mas of KCLO3 in the mixture= mas of Mixture of KClO3 and NaCl- mass of mixture remaining= 4.832-3.474=1.358 gms

Percent of KClO3= 100*1.358/4.832=28.10

Molecular weight of KClO3= 39+35.5+3*16= 122.5g/mole

Moles of KClO3= 1.358/122.5=0.0110 mole

From the reaction of 2KClO3--à 2KCl+3O2

2 moles of KClO3 gives 2 moles of KCl and 3 moles of oxygen

0.0110 mole of KClO3 gives 0.0110 mole of KCl and 1.5*0.0110mole of oxygen

Moles of oxygen lost= 0.0165. molecular weight of O2= 32

Mass of O2 lost= 32*0.0165=0.528gms

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