Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.
<u>Answer:</u> The equilibrium concentration of HCl is 
<u>Explanation:</u>
We are given:
Moles of
= 0.564 moles
Volume of vessel = 1.00 L
Molarity is calculated by using the equation:

Molarity of 
The given chemical equation follows:

<u>Initial:</u> 0.564
<u>At eqllm:</u> 0.564-x x x
The expression of
for above equation follows:
![K_c=[NH_3][HCl]](https://tex.z-dn.net/?f=K_c%3D%5BNH_3%5D%5BHCl%5D)
The concentration of pure solid and pure liquid is taken as 1.
We are given:

Putting values in above equation, we get:

Negative sign is neglected because concentration cannot be negative.
So, ![[HCl]=2.26\times 10^{-3}M](https://tex.z-dn.net/?f=%5BHCl%5D%3D2.26%5Ctimes%2010%5E%7B-3%7DM)
Hence, the equilibrium concentration of HCl is 
Answer: 150 kPa
Explanation:
Given that,
Original volume of gas V1 = 30L
Original pressure of gas P1 = 105 kPa
New pressure of gas P2 = ?
New volume of gas V2 = 21L
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
105 kPa x 30L = P2 x 21L
3150 kPa L = P2 x 21L
P2 = 3150 kPa L / 21 L
P2 = 150 kPa
Thus, 150 kPa of pressure is required to compress the gas