Answer:
okay lol, ima use this for points if okay
Explanation:
Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors.
<span> 2 Al(OH)</span>₃ → Al₂O₃ + 3 H₂O
According to equation at STP,
67.2 L (3 moles) of H₂O is produced by = 78 g of Al(OH)₃
So,
65.0 L of H₂O will be produced by = X g of Al(OH)₃
Solving for X,
X = (65.0 L × 78 g) ÷ 67.2 L
X =
75.44 g of Al(OH)₂Result: 75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry
According to that each half-life we lose half of the concentration of N2O so,
Because we start with a concentration of N2O = 0.25 mol
so after one half-life the concentration of N2O decrease to the half 0.25/2
= 12.5 x 10^-2 M
and after two half-lives the concentration of N2O of the one half-life decrease to the half (12.5 x 10^-2) / 2
=6.25 x 10^-2 M
and after three half-lives the concentration of N2O of the two half-lives decrease to the half (6.25x10^-2) / 2
= 3.1 x 10^-2 M
∴ your correct answer is 3.1 x 10^-2 M
Answer:
113.4g of N2
Explanation:
Step 1:
The balanced equation for the reaction.
N2 + 3H2 —> 2NH3
Step 2:
Determination of the mass of N2 and H2 that reacted from the balanced equation. This is illustrated below:
Molar mass of N2 = 2x14 = 28g/mol
Mass of N2 from the balanced equation = 1 x 28 = 28g
Molar mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
From the balanced equation above, 28g of N2 reacted with 6g of H2.
Step 3:
Determination of the mass N2 required to react with 24.3g of H2.
The mass of N2 required to react with 24.3g of H2 can be obtained as follow:
From the balanced equation above, 28g of N2 reacted with 6g of H2.
Therefore, Xg of N2 will react 24.3g of H2 i.e
Xg of N2 = (28 x 24.3)/6
Xg of N2 = 113.4g
Therefore, 113.4g of N2 is required to react completely with 24.3g of H2.
