Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
Answer: 
is a reactant; it is present before the reaction occurs.
Explanation:
In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.
Thus and HCl are reactants here whereas 
, 
and 
are products.
Both of them are a hope this helps
Answer:
<em>Mg = 24.30 g/mol) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Hint: 1 mole of gas at STP occupies 22.4 L</em>
40.1g of nitrogen gas is produced.
The equation given is
2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O
This equation is already balanced.
When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.
We get 1 mole of nitrogen from 3 moles of copper oxide.
We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.
4.3/3 x 1 = 1.433 mols
- 1.433 mols of nitrogen gas are produced
- The molar mass of nitrogen gas is 14+14 = 28g
- The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g
40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.
Learn more about molarity here:
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