How much kno3 should be heated to get enough oxygen required to completely burn 56l of h2?
Answers
Answer:
The equation for the decomposition of Potassium Nitrate is as follows:
2KNO3→Δ2KNO2+O2
The equation for the combustion of Hydrogen is as follows:
2H2+O2→2H2O
At STP, 1 mol of a gas occupies 22.4 L of volume. Using this as a conversion factor we can perform dimensional analysis to get the number of moles of H2 we are looking to burn. Here is the math:
5LH21⋅1.00molH222.4LH2=0.223molH2
After this, we can use the ratios from our balanced equation for the combustion of Hydrogen to figure out how many moles of Oxygen are needed. In the equation, 1 mole of Oxygen is needed for every 2 moles of Hydrogen, so we can get the number of moles of Oxygen needed by dividing the number of moles of Hydrogen we want to burn by 2. This gives us 0.223molH22=0.112molO2 .
Now that we know how much Oxygen is needed for the burn, we can use the equation for the thermal decomposition of Potassium Nitrate to figure out how much KNO3 is needed. This can be done by using the coefficients from the balanced equation we listed before. For every 2 moles of KNO3 heated, one mole of O2 is produced.
0.112molO21⋅2molKNO31molO2=0.224molKNO3
Therefore, in order to have enough Oxygen to burn 5 liters of Hydrogen gas at STP, you would need to heat 0.2 moles of Potassium Nitrate (round to 0.2 moles since “5 liters” has only one significant figure) .
Hope this helps!
Answer:
S+O
2
⟶SO
2
Moles of S =
32g/mol
16g
=0.5 mol
0.5 mol of S requires 0.5 mol of oxygen.
So, we know that sulphur burns in 0.5 mol oxygen.
The thermal decomposition of Pottasium nitrate:
2KNO
3
⟶2KNO
2
+O
2
1 mol of O
2
is produced by 2 mol of KNO
3
0.5mol of O
2
will be produced by 1 mol of KNO
3
Thus, 1 mole of KNO
3
is required to produce the given amount of oxygen.
1 mol of KNO
3
= molar mass of KNO
3
=101 g
Explanation:
it will help you