I. I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g)
a)80 gms of I2O5 reacts with 28 gm of CO.
No of moles of CO = 28/28= 1mole
No of moles of I2O5 = 80/333.84 = 0.24 moles
According to the stoichiometry, 1 mole of CO reacts with 0.2 moles of I2O5.
So,I2O5 is in excess.
According to the equation,
1mole of CO reacts with 0.2 moles of I2O5 to produce 1 mole of CO2 and 0.2 mole of I2.
Mass of iodine produced is "0.2 \u00d7 253.8 = 50.76 gm" (ans)
(b) If in the above question only 0.16 mole of I2 is produced
i) mass of I2 produced is "0.16\u00d7253.8 = 40.608 gm." (Ans)
ii)"% yield = actual yield\/ theoretical yield\u00d7 100" "%yield = actual yield\/theoretical yield\u00d7100" "Percent yield= actual\/theoretical\u00d7100"
"Percent yield =40.608\/50.76\u00d7100 =80" %(ans)
II.3 Zn(s) + 2 MoO3(s) ----------> Mo2O3(s) + 3 ZnO(s)
20 gms of MoO3 is reacted with 10 gms of Zn.
No of moles of MoO3 is "20\/143.94 = 0.139" moles
No of moles of Zn is "10\/65.38 = 0.153 moles"
Limiting reagent is Zn.
According to stoichiometry, no of moles of Zn used is equal to no of moles of ZnO produced.
No of moles of ZnO is 0.153 moles.
Mass of ZnO is "0.153\u00d781.38= 12.45 gms." (Ans)
III.2 NaOH + 2 Al + 2 H2O -----> 2 NaAlO2 + 3 H2
No of moles of NaOH required is equal to no of moles of Al required.
No of moles of Al is "51\/27 = 1.89moles"
No of moles of NaOH is "84.1\/40 = 2.1025 moles"
i)limiting reagent is Al(ans)
ii)no of moles of NaOH used is 1.89 moles
Mass of NaOH used is "1.89\u00d740=75.56gm"
Mass of NaOH not used is "84.1- 75.56 =8.54 gms" (ans)
iii)2 moles of Al is used to produce 3 moles of H2.
1 mole of Al is used to produce 1.5 moles of H2.
1.89 moles of Al is used to produce 1.5 × 1.89 moles of H2.
No of moles of H2 produced is 2.835 moles.
Mass of H2 is "2\u00d72.835 = 5.67 gms" (ans)
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