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MissTica
3 years ago
8

44.0L of O2 react with excess Sulfur dioxide gas. All gases are kept at the same temperature, pressure, and volume. What will be

the theoretical yield of Sulfur trioxide gas in Liters?
Chemistry
1 answer:
ddd [48]3 years ago
6 0

Answer:

The theoretical yield of sulfur trioxide is 88.0 Litres

Explanation:

The chemical reaction occurring between oxygen and sulfur dioxide is given in the chemical equation below :

2SO₂ + O₂ ----> 2SO₃

From the equation of reaction, 2 moles of sulfur dioxide reacts with 1 mole of oxygen to produce 2 moles of sulfur trioxide.

Volume of 1 mole of a gas at STP is 22.4 L

Number of moles of gas in 44.0 L = 44.0/22.4 moles

1 mole of oxygen reacts with 2 moles sulfur dioxide to produce 2 moles of sulfur trioxide;

44/22.4 moles of oxygen will react with excess sulfur dioxide to produce 2 × 44/22.4 moles of sulfur trioxide = 88/22.4 moles of sulfur trioxide

88/22.4 moles of sulfur trioxide will have a volume of 88/22.4 × 22.4 Litres = 88.0 Litres

Therefore, the theoretical yield of sulfur trioxide is 88.0 Litres

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What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M<br> solution?
love history [14]

Answer:

Approximately 19.1\; \rm g.

Explanation:

<h3>Number of moles of formula units of magnesium sulfate required to make the solution</h3>

The unit of concentration in this question is "\rm M". That's equivalent to "\rm mol \cdot L^{-1}" (moles per liter.) In other words:

c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}.

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L.

Calculate the number of moles of \rm MgSO_4 formula units required to make this solution:

\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}.

<h3>Mass of magnesium sulfate in the solution</h3>

Look up the relative atomic mass data of \rm Mg, \rm S, and \rm O on a modern periodic table:

  • \rm Mg: 24.305.
  • \rm S: \rm 32.06.
  • \rm O: 15.999.

Calculate the formula mass of \rm MgSO_4 using these values:

M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}.

Using this formula mass, calculate the mass of that (approximately) 0.159\; \rm mol of \rm MgSO_4 formula units:

\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}.

Therefore, the mass of \rm MgSO_4 required to make this solution would be approximately 19.1\; \rm g.

3 0
3 years ago
A sample of xenon gas collected at a pressure of 948 mm Hg and a temperature of 283 K has a mass of 128 grams. What is the volum
disa [49]

Answer:

V = 1.84 × 10³ L

Explanation:

You need to use the Ideal Gas Law and solve for volume.

PV = nRT

V = nRT/P

First, you need to convert the pressure to atm.

1 atm = 760 mm Hg

948/760 = 1.247 atm

Next, convert grams of xenon to moles.  The molar mass is 131.293 g/mol.

128/131.293 = 0.975 mol

You now have all of the values needed.

P = 1.247 atm

n = 0.975 mol

R = 8.314 J/mol*K

T = 283 K

Plug the values in and solve.

V = nRT/P

V = (0.975 × 8.314 × 283)/1.247

V = 1.84 × 10³ L

The volume of the sample will be 1.84 × 10³ L.

7 0
3 years ago
The molecular ion is not visible in the mass spectrum of 2-chloro-2- methylpropane. At what m/z value would the molecular ion be
pochemuha

Answer:  hello the complete question is attached below

Visibility of molecular ion = m/z value of 77

Explanation:

For The molecular ion to be visible, it has to be at an m/z value of 77 and this is because molecular ions will have an m/z ratio =  molecular mass of given molecule in most cases but not always in all cases.

And the visibility is possible after the removal of CH₃ ion.

ii) Evidence in the mass spectrum that suggests peak at m/z = 77

attached below

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DochEvi [55]

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