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Linolenic acid (C18H30O2 - M.W. = 278.42 g/mol) reacts with hydrogen gas according to the equation: C18H30O2 + 3H2 (g) → C18H36O

Ludmilka [50]

Answer:

2.53 L is the volume of H₂ needed

Explanation:

The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂

By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.

By stoichiometry, ratio is 1:3

Let's convert the mass of the linolenic acid to moles:

10.5 g . 1 mol / 278.42 g = 0.0377 moles

We apply a rule of three:

1 mol of linolenic acid needs 3 moles of H₂ to react

Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen

We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P

V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L

4 0

3 years ago

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

SO2(g) + NO2(g) ↔ SO3(g) + NO(g) Kc = 0.33 A reaction mixture contains 0.41 M SO2, 0.14 M NO2, 0.12 M SO3 and 0.14 M NO. Which o

stepladder [879]

Answer:

The reaction will shift in the direction of products.

Explanation:

Step 1: Data given

A reaction mixture contains:

0.41 M SO2

0.14 M NO2

0.12 M SO3

0.14 M NO

Step 2: The balanced equation

O2(g) + NO2(g) ↔ SO3(g) + NO(g) Kc = 0.33

Step 3: Define the direction of the shift of reaction:

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

Step 4: Calculate Q

Q = [NO][SO3]/[SO2][NO2]

Q = (0.14 *0.12)/(0.41*0.14)

Q = 0.0168/0.0574

Q = 0.293

Q<Kc

This means there are more reactants than products. Thud, some of the reactants will become products, causing the reaction to shift to the right.

The reaction will shift in the direction of products.

8 0

3 years ago

A mixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol

Lunna [17]

(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g) ↔ H2(g) + CO(g)

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

[H2O] [H2] [CO]

initial 0.57 + 0.15 0 0.15

change -X +X +X

Equ (0.72-X) X (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X) by solving for X

∴ X = 0.098

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X

= 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

= 0.72 - 0.098

= 0.622 M

8 0

3 years ago

Is a sauce a substance or mixture?

Komok [63]

Answer:

I would say that a sauce is a mixture, only because everything that makes up a sauce is all mixed together and turned into one thing, it isn't all separate.

5 0

3 years ago