4Fe + 3O2  2Fe2O3

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l)

Marina CMI [18]

Answer:

Please see the complete formt of the question below

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)

According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.

The answer to the above question is

The limiting reactant is the MnO₂

Explanation:

To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂

Molar mass of MnO₂ = 86.9368 g/mol

Molar mass of HCl = 36.46 g/mol

From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂

However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed

7 0

3 years ago

Identify the limiting reactant in the reaction of iron and chlorine to form FeCl3, of 22.7 g of Fe and 37.2 g of Cl2 are combine

Mice21 [21]

Answer:

3.13 g of Fe remains after the reaction is complete

Explanation:

The first step to begin is determine the reaction:

2Fe + 3Cl₂ → 2FeCl₃

Let's find out the moles of each reactant:

22.7 g / 55.85 g/mol = 0.406 moles of Fe

37.2 g / 70.9 g/mol = 0.525 moles of Cl₂

Ratio is 2:3. 2 moles of iron react with 3 moles of chlorine

Then, 0.406 moles of iron will react with (0.406 . 3)/ 2 = 0.609 moles

We need 0.609 moles of chlorine when we have 0.525 moles, so as we do not have enough Cl₂, this is the limiting reactant.

The excess is the Fe. Let's see:

3 moles of chlorine react with 2 moles of Fe

Then, 0.525 moles of Cl₂ will react with (0.525 . 2) /3 = 0.350 moles

We need 0.350 moles of Fe and we have 0.406; as there are moles of Fe which remains after the reaction is complete, it is ok that Fe is the excess reagent.

0.406 - 0.350 = 0.056 moles of Fe still remains. We convert moles to mass:

0.056 mol . 55.85g / 1 mol = 3.13 g

5 0

3 years ago

Read 2 more answers

Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57

vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

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Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0

3 years ago

216 j of energy is required to raise the temperature of aluminum from 15o to 35oc. calculate the mass of aluminum. (specific hea

kotykmax [81]

Energy is supplied to aluminium to raise the temperature of aluminium.
we can use the following equation to calculate the mass of aluminium
H = mcΔt
H - heat energy supplied
m - mass of material
c - specific heat capacity of aluminium
Δt - change in temperature - 35 °C - 15 °C = 20 °C
substituting the values in the equation
216 J = m x 0.90 J°C⁻¹g⁻¹ x 20 °C
m = 12 g
mass of aluminium is 12 g

3 0

3 years ago

In which orbitals would the valence electrons for carbon (C) be placed?

Alexxandr [17]

Answer: orbitals supernumerary

Explanation:

8 0

4 years ago