The initial volume of gas would be 8L.
initiwl volume= 9 L
using charles law for constant pressure
V1= (6/100) x 150 = 9
a.) The mass of O₂ that will be needed to burn 36.1 g B₂H₆ is 125.29 g.
b.) There will be 1.39 moles of water produced from 19.2 g of B₂H₆.
This problem is an application of stoichiometry. Stoichiometry is a topic of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data or value.
In our case, the quantitative date we want is the a.) mass of oxygen (O₂) and b.) moles of water (H₂O) both from the reaction of diborane (B₂H₆) and oxygen (O₂).
Stoichiometry problems always need a balanced chemical reaction. In the problem, the chemical reaction is not yet balanced, meaning, we still need to balance the chemical equation reaction.
Balancing the chemical reaction we have:
B₂H₆ + O₂ → HBO₂ + H₂O
B₂H₆ + 3 O₂ → 2 HBO₂ + 2 H₂O
From the balanced chemical equation there are 2 moles of water forming from 1 mole of diborane, and 3 moles of oxygen is needed to react with 1 mole of diborane. These facts are to be used in order to solve the problem.
a.) Solving for the mass of oxygen from 36.1 grams of B₂H₆, we have:
(36.1 g B₂H₆) = 125.29 grams O₂
b.) Solving for moles of water from 19.2 grams of diborane, we have:
(19.2 g B₂H₆) = 1.39 mol H₂O
Note: 27.66 g B₂H₆ is from the molar mass of diboran, and 32 g/mole is from the molar mass of oxygen; these are necessary to use in order to cancel the units.
Hence, a.) There are 125.29 grams of O₂ needed to burn 36.1 g of B₂H₆, and b.) there will be 1.39 moles of water produced from 19.2 grams of B₂H₆.
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The decrease in vapor pressure depends on the following variables:
1. the mole fraction of the amount of dissolved solute present
2. original vapor pressure of the pure solvent