The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol
according to molar ratio
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation
PV = nRT
where
P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa
V - volume
n - number of moles - 0.206 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation
136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K
V = 4.07 L
volume of SO₂ formed is 4.07 L
Answer:
Explanation:
Hello,
In this case, the described chemical reaction is:
Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:
Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:
Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:
Best regards.
Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is 22.4 L × its molar quantity, approx. 8⋅L . There are 1.51×1023molecules O2 in 10.0 g O2 .
Answer:
HF has the higher boiling point because HF molecules are more polar. Part B: CHBr3 molecules possess stronger intermolecular interaction due to higher molar mass than CHCl3
Explanation:
Fluorine is more electronegative than chlorine. This implies that HF is more polar and possess stronger hydrogen bonds than HCl molecules.
In part B, the magnitude of dispersion forces depend on molar mass, the greater the molar mass, the greater the magnitude of dispersion forces between molecules, hence CHBr3 has a greater boiling point than CHCl3
