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3 years ago

Write an equation for the reaction of hydrogen chloride and sodium sulfide to produce hydrogen sulfide with sodium chloride.

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

3 0

Answer:

sodium sulfite) Na2SO3 + HCl (hydrochloric acid). This reaction is a gas evolution reaction. The products are sodium chloride (NaCl), sulfur dioxide (SO2) and water (H2O). Sulfurous Acid (H2SO3) is produced as an intermediate in this reaction.

Explanation:

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Calculate the number of moles and the mass of the solute in each of the following solutions:

frosja888 [35]

Answer:

For a: The number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

For b: The number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

For c: The number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

For d: The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

Explanation:

To calculate the molarity of solution, we use the equation:

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

To calculate the number of moles of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(2)

For a:

Molarity of KI = $8.23\times 10^{-5}M$

Volume of solution = 325 mL = 0.325 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of KI = $2.7\times 10^{-5}mol$

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

$2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g$

Hence, the number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

For b:

Molarity of sulfuric acid = $22\times 10^{-5}M$

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

$22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of sulfuric acid = $1.65\times 10^{-5}mol$

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

$1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g$

Hence, the number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

For c:

Molarity of potassium chromate = $0.1135M$

Volume of solution = 0.250 L

Putting values in equation 1, we get:

$0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol$

Now, using equation 2, we get:

Moles of potassium chromate = $2.84\times 10^{-2}mol$

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

$2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g$

Hence, the number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

For d:

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

$3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol$

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

$39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g$

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

3 0

3 years ago

ʕ•ᴥ•ʔ Which type of reaction occurs between an acid and a base and results in the formation of a salt and water? ʕ•ᴥ•ʔ

Nataliya [291]

The answer would be D. because when an acid and base mix the start to cancel each other out causing it to neutralize

3 0

3 years ago

Read 2 more answers

a fraction collected during distillation ethanol-water what is the mass of ethanol that is present in this fraction

AURORKA [14]

Answer:

80 kg

Explanation:

because the liquid with high boiling point is going to be collected the first

4 0

1 year ago

what is the molarity of a solution prepared dissolving 317 g of CaCl2 into enough water to make 2.50 L of solution?

Minchanka [31]

Answer:

1.14 M

Explanation:

Step 1: Calculate the moles corresponding to 317 g of calcium chloride (solute)

The molar mass of calcium chloride is 110.98 g/mol.

317 g CaCl₂ × 1 mol CaCl₂/110.98 g CaCl₂ = 2.86 mol CaCl₂

Step 2: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = moles of solute / liters of solution

M = 2.86 mol / 2.50 L = 1.14 mol/L = 1.14 M

4 0

2 years ago

If you start with 10ml of 0.75 m cu(no3)2 how much cu (s) in grams should be recovered in step #7

LenaWriter [7]

0.48 grams. Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced. To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So 0.0075 mol * 63.546 g/mol = 0.476595 g. Round the results to 2 significant figures, giving 0.48 grams.

4 0

2 years ago