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Marina86 [1]
2 years ago
10

A mixture of 10cm3 of o2 and 50 cm3 of hydrogen gas sparked together to form water calculate volume of non limiting reagent left

after reacton completion
Chemistry
1 answer:
Firdavs [7]2 years ago
6 0

Answer:

30 cm³.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

O2 + 2H2 —> 2H2O

From the balanced equation above, we can say that:

1 cm³ of O2 reacted with 2 cm³ of H2 to produce 2 cm³ of H2O.

Next, we shall determine the excess reactant.

This can be obtained as follow:

From the balanced equation above,

1 cm³ of O2 reacted with 2 cm³ of H2.

Therefore, 10 cm³ of O2 will react with = (10 × 2)/1 = 10 × 2 = 20 cm³ of H2.

From the calculations made above, we can see that only 20 cm³ out of 50 cm³ of H2 given is needed to react completely with 10 cm³ of O2.

Therefore, O2 is the limiting reactant and H2 is the excess reactant (non limiting reactant).

Finally, we shall determine the volume of the non limiting reactant (excess react) that is remaining after the reaction.

This can be obtained as follow:

Volume of non limiting reactant (H2) = 50 cm³

Volume of non limiting reactant (H2) that reacted = 20 cm³

Volume of non limiting reactant (H2) remaining =.?

Volume of non limiting reactant (H2) remaining = (Volume of non limiting reactant) – (Volume of non limiting reactant that reacted)

Volume of non limiting reactant (H2) remaining = 50 – 20

= 30 cm³

Therefore, the Volume of non limiting reactant (H2) that remains after the reaction is 30 cm³

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Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you
oee [108]

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

\eta_{KClO_{3}} = \frac{m}{M}

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

\eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles                                      

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

\eta_{O_{2}} = \frac{3}{2}*0.252 moles = 0.378 moles

Finally, the mass of O₂ is:

m = 0.378 moles*32 g/mol = 12.10 g

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

6 0
2 years ago
Which chemical equation is unbalanced? c o2 right arrow. co2 sr o2 right arrow. 2sro 6h2 3o2 right arrow. 6h2o h2 h2 o2 right ar
netineya [11]

The unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical equation is one in wgich the moles of atoms in the reactants side is equal to the moles of atoms on the product side.

The given equations of reaction is not clearly stated.

Therefore, the unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

Learn more about balanced equations at: brainly.com/question/11904811

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6 0
2 years ago
The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
enot [183]

Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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2 years ago
This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det
murzikaleks [220]

Answer:

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Explanation:

<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>

Step 1: Given data

  • Formula of acetic acid: CH₃CO₂H
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Step 2: Establish the appropriate molar ratio

According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.

Step 3: Calculate the moles of atoms of hydrogen

We will use the theoretical molar ratio for acetic acid.

0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H

3 0
2 years ago
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