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2 years ago

What is the theoretical yield of NaBr

Chemistry

1 answer:

dolphi86 [110]2 years ago

7 0

The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

<h3>Balanced equation </h3>

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

<h3>How to determine the theoretical yield of NaBr</h3>

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr

Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

Learn more about stoichiometry:

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One sentence describing the difference between pure substances and mixtures.

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Answer:

A pure substance is made up of only one ingredient or combination. A mixture is made up of two or maybe more separate components that are not chemically linked.

Explanation:

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3 years ago

Read 2 more answers

Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa

Serjik [45]

Answer:

T final = 80°C

Explanation:

Q = mCpΔT

∴ Q = 18000 cal

∴ m H2O = 300 g

∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K

∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

8 0

3 years ago

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

3 years ago

Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How ma

Art [367]

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

3 0

3 years ago

How is the concentration of h3o ions and oh- ions different in an acid solution and a basic soulution?

Lubov Fominskaja [6]

In an acidic solution, the concentration of H+ is greater than the concentration of OH-. The pH will be less than 7.
In a basic solution, the concentration of OH- is greater than the concentration of H+. The pH will be greater than 7.
In a neutral solution, the concentration of H+ ions to OH-ions will be equal, and will therefore have a pH of 7. (This is due to water autoionization, which we usually ignore because it is small in other circumstances.)

5 0

3 years ago