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frosja888 [35]
3 years ago
8

At 20.°C, a 1.2-gram sample of Mg ribbon reacts rapidly with 10.0 milliliters of 1.0 M HCl(aq). Which change in conditions would

have caused
the reaction to proceed more slowly?
(1) increasing the initial temperature to 25°C
(2) decreasing the concentration of HCl(aq) to 0.1 M
(3) using 1.2 g of powdered Mg
(4) using 2.4 g of Mg ribbon
Chemistry
2 answers:
Pavlova-9 [17]3 years ago
7 0

Answer: (2) decreasing the concentration of HCl(aq) to 0.1 M

Explanation: Rate of a reaction depends on following factors:

1. Size of the solute particles: If the reactant molecules are present in smaller size, surface of particles and decreasing the size increases the surface area of the solute particles. Hence, increasing the rate of a reaction.  

2. Reactant concentration: The rate of the reaction is directly proportional to the concentration of reactants.

3. Temperature: Increasing the temperature increases the energy of the molecules and thus more molecules can react to give products and rate increases.

(1) Increasing the initial temperature to 25°C will increase the reaction rate.

(2) Decreasing the concentration of HCl(aq) to 0.1 M will decrease the reaction rate due to lesser concentration.

(3) Using 1.2 g of powdered Mg will increase the reaction rate due to large surface area.

(4) Using 2.4 g of Mg ribbon will increase the reaction rate due to high concentration of reactants.

Anuta_ua [19.1K]3 years ago
5 0
(2) is the answer. increasing temperature, increasing surface area (powdering), and increasing reactant will increase the rate of reaction
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<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

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Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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