The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.
<h3>How to calculate volume?</h3>
The volume of a gas can be calculated using the following formula:
PV = nRT
- P = pressure
- V = volume
- n = number of moles
- R = gas law constant
- T = temperature
0.75 × 0.652 = n × 0.0821 × 313
0.489 = 25.69n
n = 0.489/25.69
n = 0.019moles
Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.
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The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
Answer:
nCl= 0,062x3/2=0,093(mol)
Explanation:
Answer:
Yes
Explanation:
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