Question

I.Consider the reaction
I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g)
a) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO.
Determine the mass of Iodine, which could be produced?
b) If, in the above situation, only 0.160 moles, of Iodine was produced.
i) What mass of Iodine was produced?
ii) What percentage yield of Iodine was produced?

II. When MoO3 and Zn are heated together they react
3 Zn(s) + 2 MoO3(s) ----------> Mo2O3(s) + 3 ZnO(s)
What mass of ZnO is formed when 20.0 grams of MoO3 is reacted with 10.0 grams of Zn?

III. Aluminium dissolves in an aqueous solution of NaOH according to the following reaction:
2 NaOH + 2 Al + 2 H2O -----> 2 NaAlO2 + 3 H2
If 84.1 g of NaOH and 51.0 g of Al react:
i) Which is the limiting reagent?
ii) How much of the other reagent remains?
iii) What mass of hydrogen is produced?

Answer 1

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 $0.2 × 253.8 = 50.76 gm$ (ans)

(b) If in the above question only 0.16 mole of I2 is produced

i) mass of I2 produced is $0.16×253.8 = 40.608 gm.$ (Ans)

ii)$$ $$ $Percent yield= actual/theoretical×100$

$Percent yield =40.608/50.76×100 =80$ %(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 $0.153×81.38= 12.45 gms.$ (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 $1.89×40=75.56gm$

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 $2×2.835 = 5.67 gms$ (ans)