At one time, a common means of forming small quantities ofoxygen gas in the lab

Question

At one time, a common means of forming small quantities ofoxygen gas in the lab was to heat KClO3 2KClO3(s) -------->2KCl(s) +3O2(g)       H= -89.4kJ For this reaction, calculate H for rthe formation of(a) 1.24 mol of O2; (b) 4.89g of KCl; (c) 1.76gof KClO3 from KCl and O2. At one time, a common means of forming small quantities ofoxygen gas in the lab was to heat KClO3 2KClO3(s) -------->2KCl(s) +3O2(g)       H= -89.4kJ For this reaction, calculate H for rthe formation of(a) 1.24 mol of O2; (b) 4.89g of KCl; (c) 1.76gof KClO3 from KCl and O2.

Explanation / Answer

a)For 3 moles of O2 H=-89.4 kJ So for 1.24 moles O2 H=(-89.4/3)*1.24=-36.95kJ b)Molar wt. of KCl=74.5g 4.89g KCl=4.89/74.5=0.0656 moles KCl For 2 moles KCl H=-89.4 kJ For 0.0656 moles KCl H=(-89.4/2)*0.0656=-2.932kJ c)Molar wt of KClO3=122.5g For 245g KClO3 H=-89.4kJ For 1.76 g KClO3 H=(-89.4/245)*1.76=-0.6422kJ

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