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3 years ago

6. Barium sulfate and and sodium sulfate react in a double displacement reaction. If the

Chemistry

1 answer:

azamat3 years ago

6 0

Answer:

6) For 0.04392 moles BaSO4 we need 0.02196 moles Na2SO4

To produce 0.02196 moles Ba2SO4 and 0.04392 moles Na2SO4

7) For 16.08 moles LiOH we'll have 8.04 moles H2SO4 produced.

We need 8.04 moles Li2SO4 and 8.04 moles H2O

8) For 0.0206 moles silvernitrate we'll hace 0.0206 moles silver

9) For 0.014 moles of Cu we'll have 0.014 moles Cu(NO3)2

Explanation:

6. Barium sulfate and and sodium sulfate react in a double displacement reaction. If the reaction starts with 10.25 grams of barium sulfate what are the products and how many moles of each product is produced?

Step 1: Data given

Mass of BaSO4 = 10.25 grams

Molar mass = 233.38 g/mol

Step 2: The balanced equation

2BaSO4 + Na2SO4 → Ba2SO4 + 2NaSO4

Step 3: Calcuate moles BaSO4

Moles BaSO4 = 10.25 grams / 233.38 g/mol

Moles BaSO4 = 0.04392 moles BaSO4

Step 4: Calculate moles

For 2 moles BaSO4 we need 1 mol Na2SO4 to produce 1 mol Ba2SO4 and 2 moles NaSO4

For 0.04392 moles BaSO4 we need 0.04392/2 = 0.02196 moles Na2SO4

To produce 0.02196 moles Ba2SO4 and 0.04392 moles Na2SO4

7. Calculate the moles of Li2SO4that would be needed to produce 385 g of LiOH.

Step 1: Data given

Mass of LiOH = 385 grams

Molar mass of LiOH = 23.95 g/mol

Step 2: The balanced equation

Li2SO4 + 2H2O → 2LiOH + H2SO4

Step 3: Calculate moles LiOH

Moles LiOH = 385.0 grams / 23.95 g/mol

Moles LiOH = 16.08 moles LiOH

Step 4: Calculate moles

For 1 mol Li2SO4 we need 2 moles H2O to produce 2 moles LiOH and 1 mol H2SO4

For 16.08 moles LiOH we'll have 8.04 moles H2SO4 produced.

We need 8.04 moles Li2SO4 and 8.04 moles H2O

8. Silver nitrate reacts with copper in a single displacement reaction. To produce copper (1) nitrate and silver. If 3.50 g of silver nitrate are reacted with excess copper. How many mole of silver would be produced?

Step 1: Data given

Mass of AgNO3 = 3.50 grams

Molar mass of AgNO3 = 169.87 g/mol

Step 2: The balanced equation

Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

Step 3: Calculate moles AgNO3

Moles AgNO3 = 3.50 grams / 169.87 g/mol

Moles AgNO3 = 0.0206 moles

Step 4: Calculate moles of Ag

For 1 mol copper, we need 2 moles of silvernitrate to produce 1 mol of coppernitrate and 2 moles of silver

For 0.0206 moles silvernitrate we'll hace 0.0206 moles silver

9. (Use the chemical equation from above) How many moles of copper (1) nitrate ]can be produced with and 0.89 grams of copper metal?

Step 1: Data given

Mass of Cu = 0.89 grams

Molar mass of Cu = 63.55 g/mol

Step 2: The balanced equation

Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

Step 3: Calculate moles Cu

Moles Cu =0.89 grams / 63.55 g/mol

Moles Cu = 0.014 moles

Step 4: Calculate moles of Coppernitrate

For 1 mol copper, we need 2 moles of silvernitrate to produce 1 mol of coppernitrate and 2 moles of silver

For 0.014 moles of Cu we'll have 0.014 moles Cu(NO3)2

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Answer:

pH = 3.3

Explanation:

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Solution using the I.C.E. table:

HNO₂ ⇄ H⁺ + KNO₂⁻

C(i) 0.55M 0M 0.75M

ΔC -x +x +x

C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be

dropped giving ...

≅0.55M x ≅0.75M

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

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