5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.
Answer:
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Explanation:
it is the home button on the top left corner
not sure but h = hydrogen and o = oxygen
The chemical reaction would be:
C3H8 + 5O2 = 3CO2 + 4H2O
For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:
28.7 L C3H8 ( 1 mol / 22.41 L ) ( 4 mol H2O / 1 mol C3H8 ) ( 18.02 g / mol ) = 92.31 g H2O produced
Hope this answers the question.
Your final answer is that you will need 52.9mL of the 8.20 M of LiCl