Answer to Question #204867 in General Chemistry for Chloe Burridge

Question

Consider the unbalanced chemical reaction:  KO2 (s) + H2O(ℓ) →  KOH (s) + O2 (g) . 

If 10.70 g of KO2 and 1.80 g of H2O are reacted together:

a) What is the limiting reagent?

b) What mass of O2 can be produced?

 

c) Assuming the reaction takes place in a 1L vessel and the temperature is controlled at 25℃, what will the pressure of the vessel be?

 

c) If 2.22g of O2 is produced, what is the percent yield of the reaction?

Answer 1

A.$n_{KO_2}= \dfrac{10.70}{72}= 0.15 moles$

$n_{H_2O}= \dfrac{1.80}{18}= 0.1 moles$

Now check mole and coeffeicient ratio ,.The specy which ratio is less is the LR

In above case it is less for water hence water is the LR here.

B. Due to mole mole analysis it is clear that the moles of water is equal to the moles of moles of oxygen.

Hence moles of oxygen= 0.1 moles

weight of oxygen=$0.1 \times 32 =3.2 gm$

C.$PV=nRT$

$P= \dfrac{(0.1+0.15) \times8.314 \times298}{1}=619.4 mmHg$