The reaction is SnO2+ 2CO---->2CO2+Sn Basis : moles of SnO2= 100moles. As per th

Question

The reaction is SnO2+ 2CO---->2CO2+Sn

Basis : moles of SnO2= 100moles. As per the stoichiometry of the reaction, moles of CO= 2* moos of Sn = 2*100=200 moles

CO used is 50% excess. Moles of CO used =200*1.5= 300 moles. Since the reduction is complete, moles of CO2 formed = 200 moles, moles of Sn formed =100 moles

Products contains : CO= 300-200= 100moles, CO2=200moles and Sn= 100moles

At standard conditions, SnO2 is in solid form as well as Sn.

Hence for the reaction , SnO2(s)+ 2CO(g)------>2CO2(g)+ Sn(s),

standard heat of reaction= sum of standard heats of products- sum of standard heats of products

=2* standard heat of reaction of CO2+ 1*standard heat of reaction of Sn-{ 1* standard heat of reaction of SnO2+ 2* standard heat of reaction of Sn}, since standard heat of reaction of Sn=0

=2*(-94.05)-{1*(-138.8)+2*(-26.42)}=3.54 Kcal, the standard heat of reaction is endothermic.

Standard heat of reaction for 100moles of SnO2= 3.54*100=354 KCal

Heat of reaction = enthalpy of products+standard heat of reaction- enthalpy of reactants

Enthalpy of reactants (SnO2 and CO) : the reactants SnO2 is to be brought from 325 deg.c to 25 deg.c. Heat to be removed = moles of SnO2* specific heat* temperature difference =100*16.05*(325-25) cal=481500 cal =481.5 Kcal

Enthalpy of CO= 300*7.4*(725-25)=1554000 cal = 1554 K Cal

Total enthalpy of reactants = 481.5+1554 =2035.5 Kcal

Products contain: CO= 300-200= 100moles, CO2=200moles and Sn= 100moles. Gaseous products CO and CO2 leave at 475 deg.c

Enthalpy of CO= 100*7.4*(475-25)=333000Cal= 333 Kcal, enthalpy of CO2= 200*10.6*(475-25)=954000 Cal= 954 KCal

Tin is solid at 25 deg.c and need to be converted to liquid at 232 deg.c. This is done by supplying sensible heat to solid tin = moles of Sn* specific heat of tin* temperature difference = 100*6.5*(232-25)=134550 Cal= 134.55 Kcal

At 232 deg.c , it need to be converted to Liquid by supplying latent heat of fusion and this heat is given by= 100*1.72 Kcal= 172 Kcal

From 232 deg.c to final temperature of t, the heat to be supplied to the liquid= 100*6.6 ( specific heat of liquid tin)*(t-232)

Enthalpy of products = 333+954+134.55+172+660*(t-232)=1593.55+660*(t-232)

When there is no heat losses, deltaH= 0=enthalpy of products+ standard heat of reaction- enthalpy of reactants

=1593.55+660*(t-232)+354-2035.5 =0, t= 232.13 deg.c

Explanation / Answer

Please show clear detailed step by step work.

Metallic tin is produced by the reduction of stannic oxide (SnO2) according to the following reaction SnO2+ 2CO 2CO2+ Sn Stannic oxide crystals are fed to a furnace at 325°C. Carbon monoxide enters the furnace at 725°C and 50% in excess. The reduction is complete and the resulting tin is drawn off as a liquid. The gaseous product stream is vented from the furnace at 475°C a. For 100 mols of stannic oxide feed, determine the quantities in all other streams [20 Marks] b. Determine the standard heat of reaction and clearly state the component phases to [10 Marks] c. Assuming no heat losses, calculate the temperature at which the molten tin leaves [70 marks] which that heat of reaction relates the furnace Data Standard (25°C) heats of formation: SnO2 (s) CO (g) CO2 (g) 138.8 kcal/mol 26.42 kcal/mol 94.05 kcal/mol Heat capacities (assume constant) SnO2 (s) CO (g) CO2 (g) Sn (s) 16.5 cal/mol K 7.4 cal/mol K 10.6 cal/mol K 6.5 cal/mol K 6.6 cal/mol K Melting point of tin Heat of fusion of tin 232°C 1.72 kcal/mol Note: Tin is a solid at standard conditions. State any assumptions

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