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photoshop1234 [79]
2 years ago
11

Zinc Sulfide reacts with oxygen according to the reaction:

Chemistry
1 answer:
zhuklara [117]2 years ago
6 0

The amount in moles of the excess reactant left is 0.5 mole

<h3>Balanced equation</h3>

2ZnS (s) + 3O₂(g) --> 2ZnO (s) + 2SO₂(g)

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

<h3>How to determine the excess reactant</h3>

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

Therefore,

4.2 moles of ZnS will react with =(4.2 × 3) / 2 = 6.3 moles of O₂

From the calculations made above, we can see that only 6.3 moles of O₂ out of 6.8 moles given, is required to react completely with 4.2 moles of ZnS.

Thus, ZnS is the limiting reactant and O₂ is the excess reactant.

<h3> How to determine the mole of the excess reactant remaining</h3>

The excess reactant is O₂. Thus the mole remaining after the reaction can be obtained as illustrated below:

  • Mole of O₂ given = 6.8 moles
  • Mole of O₂ that reacted = 6.3 moles
  • Mole of O₂ remaining =?

Mole of O₂ remaining = (Mole of O₂ given) - (Mole of O₂ that reacted)

Mole of O₂ remaining = 6.8 - 6.3

Mole of O₂ remaining = 0.5 mole

Learn more about stoichiometry:

brainly.com/question/25685654

#SPJ1

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6 0
3 years ago
What gas law applies to aerosol cans being stored in a cool place?
svetlana [45]
Don't really know if this is what your asking but P1/T1= P2/T2 should show how the pressure varies with temperature (V is left out because it's constant since the gas is trapped in an aerosol can). As the temperature rises the pressure rises and if it gets too high then the can explodes, which is why it should be stored in a cool place. There's also PV=nRT might be kind of hard to find moles (n) though.


3 0
3 years ago
What is the energy change if 84.0 g of calcium oxide (CaO) reacts with excess water in the following reaction?
pochemuha
   The energy change  if 84.0 g   of CaO  react  with  excess  water is  98KJ of heat is released.

calculation
heat  =  number of moles  x  delta H

delta H = - 65.2  Kj/mol

first find the number of  moles  of  CaO reacted

moles = mass/molar mass
the molar mass  of CaO =  40 +  16=  56 g/mol
mass = 84 g
moles therefore = 84 g/56 g/mol =1.5 moles

Heat is therefore =  1.5 moles  x -65.2 = - 97.8 Kj = -98 Kj

  since  sign is  negative  the   energy  is released 

6 0
3 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
3 years ago
Describe the relationship between volume and temperature of an ideal gas
kvv77 [185]

Answer:

Explanation:

Here, we want to describe the relationship between the volume and temperature of an ideal gas

This relationship is defined by Charles' law

From this law, we know that the volume of a given mass of gas is directly proportional to its temperature at a fixed pressure

What this means is that as long as the pressure remains unchanged, when the volume increases, the temperature increases, and when the volume decreases, the temperature decreases

These can be represented by the mathematical formula below:

\frac{V_1}{T_1}\text{ = }\frac{V_2}{T_2}

8 0
1 year ago
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