HELP PLZ!CHEMISTRY may/june 2018 paper 22.
i need explanations and steps for these.
thank you so much!!

1. 12 L = 12 dm³

2. 3.18 g

Further explanation

Given

1. Reaction

K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂

69 g K₂CO₃

2. 0.03 mol/L Na₂CO₃

Required

1. volume of CO₂

2. mass Na₂CO₃

Solution

1. mol K₂CO₃(MW=138 g/mol) :

= 69 : 138

= 0.5

mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5

Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :

= 0.5 x 24 L

= 12 L

2. M Na₂CO₃ = 0.03 M

Volume = 1 L

mol Na₂CO₃ :

= M x V

= 0.03 x 1

= 0.03 moles

Mass Na₂CO₃(MW=106 g/mol) :

= mol x MW

= 0.03 x 106

= 3.18 g

Is there a diagram/picture you can provide? it seems this question goes along with a

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.