ReonRamseyz
21.11.2019 •
Chemistry
8. how many liters of oxygen gas will be produced by the decomposition of
100.0 grams of sodium nitrate? (note: 1 mole of any gas at stp will have a
vol= 22.4 liters.)
2 nano3 → 2nano2 + o2
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Ответ:
13.44 L
Explanation:
Given data:
Volume of oxygen produced = ?
Mass of sodium nitrate = 100 g
Solution:
Chemical equation:
2NaNO₃ → 2NaNO₂ + O₂
Number of moles of sodium nitrate:
Number of moles = mass/ molar mass
Number of moles = 100 g/ 85 g/mol
Number of moles = 1.2 mol
Now we will compare the moles of NaNO₃ with oxygen
NaNO₃ : O₂
2 : 1
1.2 : 1/2 × 1.2 = 0.6 mol
Volume:
one mole = 22.4 L
0.6 mol × 22.4 = 13.44 L
Ответ:
N+H→NH3
We must remember that Nitrogen and Hydrogen are both diatomic molecules in their standard gas form. This adjusts the equation to
N2+H2→NH3
Now we need to adjust coefficients in order to balance the atoms on each side of the equation. Currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.
We can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.
N2+3H2→2NH3
This gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.