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Assoli18 [71]
3 years ago
7

1 C3H8 + 5 O2 --> 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many

Chemistry
1 answer:
UNO [17]3 years ago
7 0

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen.  We also know that number of mole is proportional to pressure and volume.  Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law).  So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2.  Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol.  Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent.  Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water.  Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

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Ivanshal [37]

Answer:

red

Explanation:

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8 0
2 years ago
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Show the equation how many grams of Mg(OH)2 would be produced from 4 mol KOH?
Sindrei [870]
Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
4 moles KOH ------------------- moles Mg(OH)₂

moles Mg(OH)₂   = 4 x 1 / 2

= 2  moles of   Mg(OH)₂

molar mass Mg(OH)₂ = 58g/mol

mass of   Mg(OH)₂ = n x mm

mass of  Mg(OH)₂ = 58 x  2

= 116 g of <span> Mg(OH)₂</span>

hope this helps!
8 0
3 years ago
Compare the volume of 14.1 g of helium to 14.1 g of argon gas (under identical conditions).
s2008m [1.1K]
The Volumes can be calculated from Masses by using following Formula,

                                        Density  =  Mass / Volume
Solving for Volume,
                                        Volume  =  Mass / Density

Mass of Both Gases  =  14.1 g

Density of Argon at S.T.P  =  1.784 g/L

Density of Helium at S.T.P  =  0.179 g/L

For Argon:
                                        Volume  =  14.1 g / 1.784 g/L

                                        Volume  =  7.90 L

For Helium:
                                        Volume  =  14.1 g / 0.179 g/L

                                        Volume  =  78.77 L
4 0
3 years ago
Food dye allergies have increased in recent years. One study calculated that each 47.9 g serving of M&amp;M's candy contains 240
77julia77 [94]

Answer:

Mass percent of food dyes  = 0.0616%

Explanation:

Given data:

Mass of candy = 47.9 g

Calories = 240

Mass of fat = 10 g

Mass of carbohydrate = 34 g

Mass of protein = 2 g

Mass of food dyes = 29.5 mg

Mass percent of food dyes = ?

Solution:

First of all we will convert the mg into g.

Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g

Mass percent of food dyes  = mass of food dyes / total mass× 100

Now we will put the values.

Mass percent of food dyes  = 0.0295 g / 47.9 g × 100

Mass percent of food dyes  =  0.000616 × 100

Mass percent of food dyes  = 0.0616%

3 0
3 years ago
A gas at constant volume has a pressure of 2. 80 atm at 400. K. What will be the pressure of the gas at 360. K? 2. 52 atm 2. 94
katovenus [111]

Answer:

2.52atm for sure

Explanation:

We know that:

P1= 2.80atm

T1= 400k

T2= 360k

And we want to find P2= ?

We have an equation: P1/T1 = P2/T2

Multiple both side by T2 to remove it

P2 = 2.52

6 0
2 years ago
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