Number of moles of KCl,n= Molarity x volume in L n= 0.110 M x 0.235 L = 0.02585

Question

Number of moles of KCl,n= Molarity x volume in L

n= 0.110 M x 0.235 L = 0.02585 moles

2 moles of KCl produces 1 mole of PbCl2

0.02585 moles of KCl produces (0.02585/2)= 0.013 moles PbCl2

Mass of PbCl2,m= number of moles x molarmass

m= 0.013 mol x 177.3(g/mol)

m= 2.30 g

17) Number of moles of water,n= mass/ molarmass

n= 850/18= 47.2 mol

Number of moles of CaBr2,n'= 0.400 mol

Total moles,N= n+n'= = 47.60 mol

Molefraction of CaBr2,X= n'/N= 0.0084

18) 2 moles = 2*78 g = 156 g of benzene produces 6278 kJ of heat

M g of benzene produces 1.5*10^3 kJ

M=(156*1.5*10^3)/6278= 37.3 g

Volume,V= mass/ density

V= 37.3 g/ 0.88(g/mol)

= 42 mL

Explanation / Answer

16) According to the following reaction, what mass of PC12 can form from 235 mL ooa10M ka solution? Assume that there is excess Pb(NO3)2- 2 Ka(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq) A) 3.59 g B) 1.80 8 C)7.19 8 D) 1.308 )5.94 17) Calculate the mole fraction of CaBr2 in an aqueous solution prepared by dissolving 0.400 moles of E) 0.00841 CaBr2 in 850.0 g of water. A) 0.00900 B) 0.0167 C) 0.0252 D) 0.0540 18) What volume of benzene (CoHó d 0.88 g/mL, molar mass 78.11 g/mol) is required to produce 1.5 103 kJ of heat according to the following reaction? H.xn--6278k] D) 19 mL 2C6H6+1502(g) 12 CO2(g) + 6H20(g) C) 21 mL E)37 mL A) 75 mL B) 42 mL

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