What amount of heat, in kJ, is required to vaporize 163.45 g of ethanol (C₂H₅OH)? (∆Hvap = 43.3 kJ/mol)
1 answer:
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Answer:
2.85moles of oxygen gas
Explanation:
Given parameters:
Volume of oxygen gas = 63.8L
Unknown:
Number of moles = ?
Solution:
We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:
1 mole of a gas at STP occupies a volume of 22.4L
So;
63.8L of oxygen gas will take up a volume of
= 2.85moles of oxygen gas
Answer:
The molar mass of CO2 = 44.01 g/mol
Explanation:
Start with 2.09 moles of HCl and convert to moles of CO2 For every 2 moles of HCl used, 1 mole of CO2 is formed.
2.09 mol HCl •(1 mol CO2/2 mol HCl) = 1.045 mol CO2
Now convert moles of CO2 to grams using the molar mass.
1.045 mol CO2 • 44.01 g/mol = 45.99 grams CO2
3 significant figures = 46.0 grams CO2
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