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emilycabrera610
17.01.2021 •
Chemistry
HELP PLZ!CHEMISTRY
may/june 2018 paper 22.
i need explanations and steps for these.
thank you so much!!
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Ответ:
1. 12 L = 12 dm³
2. 3.18 g
Further explanationGiven
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
Ответ:
B
Explanation: