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3 years ago

How many grams of oxygen are required too burn 8.8g of c3h8?

1 answer:

4 0

From the balanced equation:
1mol C3H8 requires 5mol O2 for combustion 
Molar mass C3H8 = 44g/mol 
8.8g C3H8 = 8.8/44 = 0.2mol C3H8 
This will require 5*0.2 = 1.0mol O2 
Molar mass O2 = 32g/mol 
Therefore 32g of O2 required.

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Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago

Are usually solid shiny and conductive object

crimeas [40]

If the solid is a metal, yes. Most solids on the periodic table are metals

8 0

4 years ago

At a room temperature, which of the following will evaporate first, a drop of water or a drop of ethanol

Angelina_Jolie [31]

Ethanol will evaporate first because it is more volatile or has a lower boiling point than water.

5 0

3 years ago

Read 2 more answers

Why do lithium (Li) and sodium (Na) have similar properties?

ella [17]

B, they belong to group 1

A: Li is period 2 and Na is period 3

C: Na has one more electron shell

D: Not related to how it is spelled

8 0

3 years ago

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

Simora [160]

Answer:- Density is 10.5 grams per mL and the substance is Silver(Ag).

Solution:- Mass of jewelry is given as 132.6 g. Initial volume of water in the graduated cylinder is 48.6 mL and the volume becomes 61.2 mL when this jewelry is submerged in the cylinder.

So, the volume of the jewelry = 61.2 mL - 48.6mL = 12.6 mL

We know that, $density=\frac{mass}{volume}$

Let's plug in the values in the above formula to calculate the density:

$density=\frac{132.6g}{12.6mL}$

$density=10.5\frac{g}{mL}$

So, the density of the jewelry is $10.5\frac{g}{mL}$ .

The substance from which the jewelry is made could easily be identified by comparing the calculated density with the density values given for different substances in the density table.

Looking at the density table, 10.5 grams per mL is the density for silver. So, the jewelry is made up from Silver.

4 0

3 years ago