We have to solve this question using the stoichiometry of the reaction:
The equation of the reaction is;

According to the question;
Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles
From the stoichiometry of the reaction:
Since;
24 moles of CO2 released 15,026 KJ
0.48 moles of CO2 will release 0.48 * 15,026/24
= 301 KJ of heat.
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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.
C- more than one light year or B-exactly one light year
1.34 L of HF
Explanation:
We have the following chemical reaction:
Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)
First we calculate the number of moles of SnF₂:
number of moles = mass / molecular weight
number of moles of SnF₂ = 5 / 157 = 0.03 moles
From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.
Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:
number of moles = volume / 22.4 (L/mole)
volume of HF = number of moles × 22.4
volume of HF = 0.06 × 22.4 = 1.34 L
Learn more about:
problems with gases at STP
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