mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
Answers:
(a) 30.55 °C
(b) 298 K and 77°F
(c) 204.44 °C and 477.44 K
(d) -320.8 °F and -196 °C
Explanation:
Converting °C into °F;
°F = °C × 1.8 + 32
Converting °F into °C;
°C = °F - 32 ÷ 1,8
Converting °C into K;
K = °C + 273
Converting K into °C;
°C = K - 273
Answer is: volume will be 3.97 liters.
Boyle's Law: the pressure volume law - volume of a given amount of gas held varies inversely with the applied pressure when the temperature and mass are constant.
p₁V₁ = p₂V₂.
p₁ = 755 torr.
V₁ = 5.00 l.
p₂ = 1.25 atm · 760 torr/atm.
p₂ = 950 torr.
755 torr · 5 l = 950 torr · V₂.
V₂ = 755 torr · 5 l / 950 torr.
V₂ = 3.97 l.
When pressure goes up, volume goes down.
When volume goes up, pressure goes down.
