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boyakko [2]
2 years ago
15

Please help 100 points!

Chemistry
2 answers:
elena-s [515]2 years ago
7 0

Explanation:

The balance chemical equation is

2H2+O2—>2H2O

The limiting reactant is found by dividing the moles of each reactant by its coefficient

For H2=10.39/2=5.195

For O2=16.17/1= 16.17

Since H2 has the smaller value it’s the limiting reactant

The mass of the maximum amount of H2O is found by using the formula

moles= mass/molar mass

solving for mass gives

mass= moles*molar mass

(molar mass of H2O= 18g/mol)

mass= 5.195*18=93.51grams of H2O

zzz [600]2 years ago
3 0

Balanced chemical equation: 2H2 + O2 → 2H2O

The limiting reagent will be H2

All 10.39 mol of H2 will be used up produce 10.39 mol of H20

The excess reagent will be O2 and there will be 5.78 mols left over

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When an acid and base react, salt and water is formed.<br><br> True<br><br> False
lora16 [44]

Answer:

im pretty sure its true.

Explanation:

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3 years ago
A chemist titrates of a hypochlorous acid solution with solution at . Calculate the pH at equivalence. The of hypochlorous acid
const2013 [10]

The question is incomplete, here is the complete question:

A chemist titrates 110.0 mL of a 0.2412 M hypochlorous acid (HCIO) solution with 0.0613 M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of hypochlorous acid is 7.50. Round your answer to 2 decimal places

<u>Answer:</u> The pH of the solution is 10.09

<u>Explanation:</u>

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HClO

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=0.2412M\\V_1=110.0mL\\n_2=1\\M_2=0.0613M\\V_2=?mL

Putting values in above equation, we get:

1\times 0.2412\times 110.0=1\times 0.0613\times V_2\\\\V_2=\frac{1\times 0.2412\times 110.0}{1\times 0.0613}=432.8mL

At equivalence, the number of moles of acid is equal to the number of moles of base. Also, the moles of salt which is NaClO will also be the same.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For HClO:</u>

Molarity of HClO solution = 0.2412 M

Volume of solution = 110.0 mL

Putting values in equation 1, we get:

0.2412M=\frac{\text{Moles of HClO}\times 1000}{110}\\\\\text{Moles of HClO}=\frac{(0.2412\times 110)}{1000}=0.026532mol

  • <u>For NaClO:</u>

Moles of NaClO = 0.026532 moles

Volume of solution = [432.8 + 110] mL = 542.8 mL

Putting values in above equation, we get:

\text{Molarity of NaClO}=\frac{0.026532\times 1000}{542.8}=0.0489M

To calculate the pH of the solution, we use the equation:

pH=7+\frac{1}{2}[pK_a+\log C]

where,

pK_a = negative logarithm of weak acid which is hypochlorous acid = 7.50

C = concentration of the salt = 0.0489 M

Putting values in above equation, we get:

pH=7+\frac{1}{2}[7.50+\log (0.0489)]\\\\pH=7+3.09=10.09

Hence, the pH of the solution is 10.09

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