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vladimir1956 [14]
3 years ago
12

Given the following rate equation, complete the sentences to identify the reaction order for each reagent, the overall reaction

order, and what effect doubling substance A and halving substance B would have on the overall rate of the process.
Rate= k[A] [B]^3

a. zero
b. first
c. second
d. third
e. fourth
f. remain the same as
g. be 1/4 of
h. be 1/2 of
i. be twice
j. be four times


1. Given the rate equation. the reaction order with respect to A is _________order
2. Given the rate equation, the reaction order with respect to B is _________ order
3. Given the rate equation. the overall reaction order I s________ order
4. Given the rate equation. when the concentration of A is doubled while that of B is halved, the rate will ________the original rate.
Chemistry
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

1. First

2. Third

3. Fourth

4.remain the same as

Explanation:

Given the reaction equation;

Rate= k[A] [B]^3

We can see that the order of reaction is first order with respect to reactant A and third order with respect to reactant B. This gives an overall fourth order reaction.

If the concentration of A is doubled and that of B is halved. The rate of reaction remains the same.

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 % (m/m) = 2% = 2/100

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agasfer [191]

Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

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