All categories

3 years ago

2Fe +3S= Fe2S3 how many grams of sulfur are involved in this reaction if 95 grams of iron are needed to react with sulfur

Chemistry

1 answer:

Zepler [3.9K]3 years ago

5 0

Answer:

83 g

Explanation:

Step 1: Write the balanced equation

2 Fe + 3 S ⇒ Fe₂S₃

Step 2: Calculate the moles corresponding to 95 g of Fe

The molar mass of Fe is 55.85 g/mol.

95 g × 1 mol/55.85 g = 1.7 mol

Step 3: Calculate the moles of S needed to react with 1.7 moles of Fe

The molar ratio of Fe to S is 2:3. The moles of S needed are 3/2 × 1.7 mol = 2.6 mol

Step 4: Calculate the mass corresponding to 2.6 moles of S

The molar mass of S is 32.07 g/mol.

2.6 mol × 32.07 g/mol = 83 g

You might be interested in

What are function of lipids

olga2289 [7]

Lipids serve as structural components of cell membranes, function as energy storehouses, and function as important signalling molecules. They include fats, waxes, oils, hormones, and certain components of membranes and function as energy-storage molecules and chemical messengers.

Hope this helped :)

6 0

2 years ago

Tectonic plates make up which of the following layers of the Earth?

forsale [732]

The lithosphere makes up the layer of the earth.

6 0

4 years ago

guys please help me WILL MARK BARINLIEST If you have 5.7 x 10^24 molecules of O2 at STP, how many liters of O2 do you have? must

zmey [24]

Answer:

212.8 dm^3 or L

Explanation:

1 mole of any sub=6.02×10^23 molecules

X mole of O2=5.7×10^24 molecules

X mole=5.7×10^24/6.02×10^23

=9.5 mole

1 mole of any gas at stp=22.4 dm^3

Therefore, 9.5 mole of O2 will be 22.4×9.5

=212.8 dm^3

5 0

3 years ago

How many molecules of co2 in 97.3 grams of co2

s2008m [1.1K]

Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Reading a Periodic Table
Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

97.3 g CO₂

Step 2: Define conversions

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$97.3 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.33138 × 10²⁴ molecules CO₂

Step 4: Check

We are given 3 sig figs. Follow sig fig rules.

1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂

8 0

3 years ago

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = 0,0989 moles NaOH

Moles of Al₂(SO₄)₃ are:

15,8g × $\frac{1mol}{342,15g}$ = 0,0462 moles Al₂(SO₄)₃

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × $\frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH}$ = 0,0165 moles Al₂(SO₄)₃

As you have 0,0462 moles Al₂(SO₄)₃ the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × $\frac{2molAl(OH)_{3}}{6 moles NaOH}$ = 0,0330 moles of Al(OH)₃

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = 2,57 g of Al(OH)₃ ≡ mass of precipitate

I hope it helps!

3 0

3 years ago