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RoseWind [281]
3 years ago
6

How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2

Chemistry
2 answers:
NNADVOKAT [17]3 years ago
6 0

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂     →     CO₂

Number of moles of  CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

                          CO₂         :       O₂  

                              1           :         1

                            0.84       :       0.84

0.84 moles of oxygen are required.

dezoksy [38]3 years ago
6 0
I solved the problem and this is correct

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80.0 mL of 0.30 M NaOH and 80.0 mL of 0.30 M HCl are mixed together. What is the approximate pH
Licemer1 [7]

Answer:

  • <u><em>Neutral</em></u>

Explanation:

The number of moles of solute is equal to product of the molar concentration (molarity) and the volume (in liters) of solution.

Since the volumes and the molar concentrations of the<em> NaOH </em>and <em>HCl </em>solutions mixed are equal, each one of them contributes the same number of moles of solute.

Since every mol of NaOH produces one mol of OH⁻ ions and every mol of HCl produces one mol of H⁺ ion, the number of moles of OH ⁻ and H⁺ in solution are equal.

Thus, OH⁻ and H⁺ ions will be neutralized by the reaction:

  • OH⁻ (aq) + H⁺ (aq) ⇄ H₂O (l)

Which is strongly shifted to the right and has <em>neutral pH</em>.

Hence, you conclude that the approximate <em>pH of the solution is neutral.</em>

6 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
In the following reaction, how many liters of oxygen produce 560 liters of
saw5 [17]

The balanced chemical reaction will be:

CH4 + 2O2 → CO2 + 2H2O

We are given the amount of carbon dioxide to produce from the reaction. This will be our starting point.

 

560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) ( 22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>

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3 years ago
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