Answer:
V = 10.3 L
Explanation:
Given data:
Mass of methane = 6.40 g
Volume of CO₂ produced = ?
Temperature = 35°C (35+273 = 308 K)
Pressure = 100.0 KPa (100.0/101 = 0.98 atm)
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 6.40 g/ 16 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of CO₂ with CH₄.
CH₄ : CO₂
1 : 1
0.4 : 0.4
Volume of CO₂:
Formula:
PV = nRT
0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K × 308 K
0.98 atm ×V = 10.11 atm.L
V = 10.11 atm.L /0.98 atm
V = 10.3 L
As we know that neutralization reaction is a reaction in which base react with acid to form salt and water.
When Potassium Hydroxide reacts with Sulphuric Acid, it forms Potassium Sulphate and Water.
As a result of neutralization reaction, Potassium Sulphate and Water is formed.
2KOH + H2SO4 ----> K2SO4 + 2H2O
Here, K2SO4 is found in aqueous medium in neutralization reaction. It is a neutral salt.
It would be Rubidium chloride.
Answer:
C
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