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<span>2Al + 3Br2 -------------> 2AlBr3
</span>3 g Al = 0.11 mol Al.
<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>
<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>
<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>
</span>
Answer;
4.5 m³
Solution:
The statement says that two blocks are present on a lid of a container with volume of 9 m³. The mass of lid is equal to the mass of two blocks. It means that initially there are four blocks (or four atm pressure) upon 9 m³ volume.
After that four more blocks are placed on the lid. Means the pressure is increased from 4 atm to 8 atm (2 atm of lid, 2 atm of old blocks, 4 atm of new four blocks).
So, Data generated is,
P₁ = 4 atm
V₁ = 9 m³
P₂ = 8 atm
V₂ = ?
According to Boyle's Law,
P₁ V₁ = P₂ V₂
Solving for V₂,
V₂ = P₁ V₁ / P₂
Putting values,
V₂ = (4 atm × 9 m³) ÷ 8 atm
V₂ = 4.5 m³
