Answer:
a. the mole fraction of CO in the mixture of CO and O2.
mole fraction = moles of CO/ Total moles of the mixture
Mole fraction of CO = 10/(10+12.5)=0.444
b. Reaction - CO(g)+½O2(g)→CO2(g)
Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2
So given,
At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.
3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2
This means that unused mols are : 7mols of CO and 11mols of O2
Total product mixture = 3 + 7 + 11 = 21mols
mole fraction of CO = 7/21 = 0.33
One of the most likely products for the reaction would be
<h3>Chemical reactions</h3>
The reaction between and yields 3 products which are (a precipitate), , and as shown by the equation below:
One of the products precipitates out of the solution to give the reaction a precipitation reaction look.
More on precipitation reaction can be found here: brainly.com/question/24158764
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Answer: 10.2 grams
Explanation:
The balanced chemical reaction is :
According to the ideal gas equation:
P = Pressure of the gas = 740 torr = 0.97 atm (760torr=1atm)
V= Volume of the gas = 12.0 L
T= Temperature of the gas = 19°C = 292 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas
According to stoichiometry:
2 moles of hydrogen are generated by = 1 mole of
Thus 0.48 moles of hydrogen are generated by = moles of
Mass of
Thus 10.2 grams of are needed to generate 12.0 L of hydrogen gas if the pressure of hydrogen is 740. torr at 19°C