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Ivan
3 years ago
13

How many grams of sulfuric acid are needed to produce 57.0 grams of water? Show all steps of your calculation as well as the fin

al answer.
2NaOH+H2SO4 > 2Na2SO4+2H2O
Chemistry
2 answers:
antiseptic1488 [7]3 years ago
6 0

Answer:

155.16 g.

Explanation:

  • Firstly, It is considered as a stichiometry problem.
  • From the balanced equation: 2NaOH + H₂SO₄ → 2Na₂SO₄ + 2H₂O
  • It is clear that the stichiometry shows that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to give 2.0 moles of Na₂SO₄ and 2.0 moles of H₂O.
  • We must convert the grams of water (57.0 g) to moles <em>(n = mass/molar mass)</em>.
  • n = (57.0 g) / (18.0 g/mole) = 3.1666 moles.
  • Now, we can get the number of moles of H₂SO₄ that is needed to produce 3.1666 moles of water.
  • <em>Using cross multiplication:</em>
  • 1.0 mole of H₂SO₄ → 2.0 moles of H₂O, <em>from the stichiometry of the balanced equation</em>.
  • ??? moles of H₂SO₄ → 3.1666 moles of H₂O.
  • The number of moles of H₂SO₄ that will produce 3.1666 moles of H₂O <em>(57.0 g)</em> is (1.0 x 3.1666 / 2.0) = 1.5833 moles.
  • Finally, we should convert the number of moles of H₂SO₄ into grams <em>(n = mass/molar mass)</em>.
  • Molar mass of H₂SO₄ = 98.0 g/mole.
  • mass = n x molar mass = (1.5833 x 98.0) = 155.16 g.
Sergeu [11.5K]3 years ago
5 0

<u>Answer:</u> 155.134 grams of sulfuric acid is needed.

<u>Explanation:</u>

To calculate the moles, we use the following equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of water:

Given mass of water = 57 grams

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

\text{Number of moles}=\frac{57g}{18g/mol}=3.166moles

For the given chemical reaction, the equation follows:

2NaOH+H_2SO_4\rightarrow 2Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

2 moles of water are produced by 1 mole of sulfuric acid

So, 3.166 moles of water will produced by = \frac{1}{2}\times 3.166=1.583moles of sulfuric acid.

Now, to calculate the mass of sulfuric acid, we use the moles equation:

Molar mass of sulfuric acid = 98 g/mol

Putting values in above equation, we get:

1.583mol=\frac{\text{Given mass}}{98g/mol}

Mass of sulfuric acid = 155.134 grams

Hence, 155.134 grams of sulfuric acid is needed.

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The balanced redox reaction is

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

<h3>What is Galvanic cell? </h3>

It is a device that is used for the conversion of the chemical energy which is produce in the redox reaction into the electrical energy. It is also termed as the voltaic cell or electrochemical cell.

In the voltaic cell, at anode which is negative electrode of voltaic cell oxidation occurs and at the cathode which is positive electrode, reduction occurs

<h3>What is redox reaction? </h3>

The reaction in which both reduction and oxidation reaction takes place simultaneously is known as redox reaction.

<h3>What is oxidation reaction? </h3>

The reaction in which a substance or compound or species looses its electrons. In this reaction, oxidation state of an element increases.

<h3>What is reduction reaction? </h3>

The reaction in which a substance or compound or species accept the electrons. In this reaction, oxidation state of an element decreases.

The given two-half reactions are:

At anode:

3Fe → 3Fe²⁺ + 6e-

At cathode:

8H(+) + 2NO₃⁻ → 2NO + 4H2O

Redox reaction:

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

According to the voltaic cell, the half-cell reaction (1) shows oxidation reaction therefore, it occurs at anode and the half-cell reaction (2) shows reduction reaction therefore, it occurs at cathode.

Thus, we concluded that the redox reaction is

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

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2 years ago
Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
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Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of Ag_2O = 25.0 g

Mass of C_{10}H_{10}N_4SO_2 = 50.0 g

Molar mass of Ag_2O = 231.7 g/mole

Molar mass of C_{10}H_{10}N_4SO_2 = 250.3 g/mole

Molar mass of AgC_{10}H_{9}N_4SO_2 = 357.1 g/mole

First we have to calculate the moles of Ag_2O and C_{10}H_{10}N_4SO_2.

\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles

\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)

From the balanced reaction we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 1 mole of Ag_2O

So, 0.1998 moles of C_{10}H_{10}N_4SO_2 react with \frac{0.1998}{2}=0.0999 moles of Ag_2O

From this we conclude that, Ag_2O is an excess reagent because the given moles are greater than the required moles and C_{10}H_{10}N_4SO_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgC_{10}H_9N_4SO_2

From the reaction, we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 2 mole of AgC_{10}H_9N_4SO_2

So, 0.1998 mole of C_{10}H_{10}N_4SO_2 react with 0.1998 mole of AgC_{10}H_9N_4SO_2

Now we have to calculate the mass of AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

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Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
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Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

                    NaOH    +     HCl ⇒ NaCl + H₂O

Initial           0.0200       0.0188

Reaction    -0.0188       -0.0188

Final         1.20 × 10⁻³          0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³  moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

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