It’s the first 1 M yea it’s the first one
Answer:
Mass of solution=100g
mass of salt=20g
so; mass of solute=80g
percentage composition =(mass of salt/total
mass) ×100
= \frac{20}{100} \times 100 \\ = 20\%
glad to help you
hope it helps
__ KClO₃ → __ KCl + __ O₂
Left Side:
1 K
1 Cl
3 O
Right Side:
1 K
1 Cl
2 O
Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.
This gives us 2KClO₃ → __ KCl + 3O₂.
However, this equation is still not balanced.
Left Side:
2 K
2 Cl
6 O
Right Side:
1 K
1 Cl
6 O
In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.
2KClO₃ → 2KCl + 3O₂
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
