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3 years ago

Using this balanced equation: 2 NaOH + H2SO4 —> H2O + Na2SO4

Chemistry

1 answer:

mars1129 [50]3 years ago

7 0

<h3>Answer:</h3>

266.325 g

<h3>Explanation:</h3>

We are given the balanced equation;

2NaOH + H₂SO₄ → H₂O + Na₂SO₄

Mass of NaOH as 150 g

We are required to determine the mass of Na₂SO₄ that will be formed.

<h3>Step 1: Determine the number of moles of NaOH</h3>

Moles = Mass ÷ molar mass

Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 150 g ÷ 40 g/mol

= 3.75 moles

<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>

From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1

Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2

= 3.75 moles ÷ 2

= 1.875 moles

<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>

we know that;

Mass = Moles × Molar mass

Molar mass of Na₂SO₄ is 142.04 g/mol

Therefore;

Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol

= 266.325 g

Thus, the mass of sodium sulfate formed 266.325 g

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Suppose that 25.0 mL of 0.440 M sodium chloride is added to 25.0 mL of 0.320 M silver nitrate. How many moles of silver chloride

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The number of moles of silver chloride that will precipitate is 0.008 mole

From the question,

We are to determine the number of moles of silver chloride that will precipitate

First,

We will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

NaCl + AgNO₃ → AgCl + NaNO₃

This means

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride and 1 mole of sodium nitrate

Now, we will determine the number of moles of each reactant present

For sodium chloride (NaCl)

Concentration = 0.440 M

Volume = 25.0 mL = 0.025 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of NaCl present = 0.440 × 0.025

Number of moles of NaCl present = 0.011 mole

For silver nitrate (NaNO₃)

Concentration = 0.320 M

Volume = 25.0 mL = 0.025 L

∴ Number of moles of NaNO₃ present = 0.320 × 0.025

Number of moles of NaNO₃ present = 0.008 mole

From the balanced chemical equation,

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride

Then,

0.008 mole of sodium chloride will react with the 0.008 mole of silver nitrate to produce 0.008 mole of silver chloride

∴ 0.008 mole of silver chloride will be produced

Hence, the number of moles of silver chloride that will precipitate is 0.008 mole

Learn more here: brainly.com/question/18434602

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2 years ago

The elemental analysis of an organic solid extracted from gum arabic (a gummy substance used in adhesives, inks, and pharmaceuti

Evgesh-ka [11]

Answer:

The molar mass of the organic solid is 120.16 g/mol.

The molecular formula of an organic solid is $C_4H_8O_4$

Explanation:

Let the molecular mass of an organic solid be $C_xH_yO_z$

$\Delta T_b=K_b\times m$

$Delta T_b=K_b\times \frac{\text{Mass of solid}}{\text{Molar mass of solid}\times \text{Mass of diphenyl in Kg}}$

where,

$\Delta T_f$ =Elevation in boiling point = $=1.5^oC$

Mass of organic solid= 0.561 g

Mass of diphenyl = 24.9 g = 0.0249 kg (1 kg = 1000 g)

$K_b$ = boiling point constant = 8.00 °C/m

m = molality

Now put all the given values in this formula, we get

$1.5^oC=8.00 ^oC/m\times \frac{0.561 g}{\text{Molar mass of solid}\times 0.0249 kg}$

${\text{Molar mass of solid}}=120.16 g/mol$

$\%=\frac{\text{Number of atoms}\times \text{mass of an atom}}{\text{molas mass of compound}}\times 100$

Percentage of carbon in an organic solid = 40.0%

$40\%=\frac{x\times 12 g/mol}{120.16 g/mol}\times 100$

x = 4.0

Percentage of hydrogen in an organic solid = 6.7%

$6.7\%=\frac{y\times 1 g/mol}{120.16 g/mol}\times 100$

y = 8.0

Percentage of hydrogen in an organic solid = 6.7%

$53.3\%=\frac{x\times 12 g/mol}{120.16 g/mol}\times 100$

y = 4.0

The molecular formula of an organic solid is $C_4H_8O_4$

8 0

3 years ago