Answer:
So the volume will be 2.33 L
Explanation:
The reaction for the combustion is:
2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)
mass of butane to moles (mass / molar mass)
1.4 g / 58 g/mol
= 0.024 moles
2 moles of butane can produce 8 moles of carbon dioxide
0.024 moles of butane must produce (0.024 × 8) /2
= 0.096 moles of CO₂
Now we apply the Ideal Gases Law to find out the volume formed.
P . V = n . R . T
p = 1atm
n = 0.096 mol
R = 0.082 L.atm/mol.K
T = 273 + 23 = 296K
V = ?
1atm × V = 0.096 mol × 0.082 L.atm/mol.K × 296K
V = 0.096 mol × 0.082 L.atm/mol.K × 296K / 1atm
= 2.33 L
So the volume will be 2.33 L
Answer:
2Mg + O2 → 2MgO
Explanation:
In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.
Solution. because the substances can't go back to their original form. like kool aid, when its mixed you cant separate the powder and the water again.
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
1 hectoliter is 26.4172
1 kiloliter is 264.172
