Container A holds 717 mL of ideal gas at 2.50 bar . Container B holds 179 mL of ideal gas at 4.30 bar . If the gases are allowed
1 answer:
Answer:
Pressure gas A
using boyles law
=
V2
V2 = 717ml + 179 ml
= 896ml
∴ = 2.50 × 717ml/896ml
= 2.0 bar
Pressure B
P2 = 4.30 bar× 179ml/896ml
= 0.859bar
ptotal = +
= 2.0 bar + 0.859 bar
= 2.859 bar
Explanation: Using Daltons law of partial pressure,the pressure is independently of each other when the gas is exerted.where we can use daltons law to find the pressure of each gas separately when it expands into the total volume in two containers.
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First a balanced reaction equation must be established:
→ 
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.