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IgorC [24]
3 years ago
12

Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make ___ mol ammonium sulfate

when you neutralize 11.00 g ammonium hydroxide. __ NH4OH + __ H2SO4 → __ (NH4)2SO4 + __ H2O
Chemistry
1 answer:
Bingel [31]3 years ago
3 0

Answer:

1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • We can balance the equation by applying the conservation of mass to the equation.
  • The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
  • This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
  • We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass. </em>

n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.

<u><em>Using cross multiplication: </em></u>

2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.

0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.

∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.

<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>

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3 years ago
At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.
Goryan [66]

Answer:

Kc = 2.34 mol*L

Explanation:

The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.

A + B ⇄ C + D

Kc = [C] * [D] / [A] * [B]

According to the reaction

Kc = [SO2]^2 * [O2]^2 / [SO3]^2

Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3

0.450 --> 0 + 0 (Beginning of the reaction)

0.260 --> 0.260 + 0.130 (During the reaction)

0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)

Kc = [0.260]^2 + [0.130]^2 / [0.190]^2

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3 0
3 years ago
A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55
Xelga [282]

<u>Answer:</u> The volume of NaOH required is 402.9 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For HCl:</u>

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol

  • <u>For sulfuric acid:</u>

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL

Hence, the volume of NaOH required is 402.9 mL

6 0
3 years ago
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