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Vinil7 [7]
3 years ago
14

How do you write the balanced equation ​

Chemistry
1 answer:
katrin [286]3 years ago
6 0

Answer:

See BELOW!

Explanation:

Your question is confusing, you are not mentioning specifically what you are trying to balance the equation of. If you require more assistance, write in the comments and I'll be glad to assist. you!

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Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose
Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624  = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

3 0
3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0
3 years ago
If a 6 ohm wire is connected to a 10 volt battery, what will the current be?
lidiya [134]

Answer:

60 amperes

Explanation:

5 0
3 years ago
Read 2 more answers
A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t
andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

              = 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0
3 years ago
The chemical equation below shows the combustion of methane (CH4).
pishuonlain [190]
According to the chemical equation, the reaction ratio between O2 and CO2 is 2:1, which mean for every 2 moles of O2 reacted there is 1 mole of CO2 formed. 

Use the molar mass and mass of O2 to find out the moles of O2: moles of O2 = mass of O2/molar mass of O2 = 8.94g/32.00g/mol = 0.2794 mole.  Therefore, the moles of CO2 that formed is 0.2794moles/2 = 0.1397 mole

Use the moles and molar mass CO2 to find out the mass of CO2:
Mass of CO2 = moles of CO2 * molar mass of CO2 = 0.1397 mole * .44.01g/mole = 6.15 g.

So the answer is B 6.15g.
3 0
3 years ago
Read 2 more answers
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