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Vesnalui [34]
3 years ago
14

When 4.15 grams of silver nitrate is reacted with 1.11 grams of iron(III) chloride, which best represents the amount of silver c

hloride produced?
Chemistry
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

The mass of silver chloride produced = 2.202 g

Explanation:

Equation of the reaction is given below

3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)

molar mass of AgNO₃ = 170 g/mol

molar mass of FeCl₃ = 233.5 g/mol

molar mass of AgCl = 143.5 g/mol

3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate

4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃

1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃

mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1

therefore, FeCl₃ is the limiting reactant

0.0047 moles of FeCl₃ reacting will produce 0.0047 *  3 moles of AgCl = 0.0141 moles of AgCl

0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =

Therefore mass of silver chloride produced = 2.202 g

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The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00
Marina CMI [18]

<u>Answer:</u> The moles of CO_2 added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of CO_2 at equilibrium = 1.00 moles

Moles of H_2 at equilibrium = 1.00 moles

Moles of H_2O at equilibrium = 2.40 moles

Moles of CO at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

The given chemical equation follows:

CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)

The expression of K_c for above equation follows:

K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}

Putting values in above equation, we get:

K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76

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)}}

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of CO_2 needed be 'x' moles.

Now, equilibrium gets re-established:

              CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)

Initial:       1.00      1.00              2.40       2.40

At eqllm:   (0.236+x)   0.236      3.164     3.164

Again, putting the values in the expression of K_c, we get:

5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13

Hence, the moles of CO_2 added to the system is 7.13 moles

4 0
2 years ago
Which of the following is a correct statement regarding the free energy?
Mnenie [13.5K]

Answer:

A

Explanation:

6 0
2 years ago
A 520-gram sample of seawater contains 0.317 moles of NaCl. What is the percent composition of NaCl in the water?
Volgvan

Answer:

c

Explanation:

8 0
3 years ago
14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
Murrr4er [49]

Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
3 years ago
4 examples of bases​
Vesna [10]

Answer:

Caustic Soda

Limewater

Milk of Magnesia

Bleach

Explanation:

4 0
2 years ago
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