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2 years ago

For the reaction So3 + H2O -> H2SO4, how many grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid

Chemistry

1 answer:

zalisa [80]2 years ago

3 0

Answer:

320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.

Explanation:

The balanced reaction is:

SO₃ + H₂O → H₂SO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

SO₃: 1 mole
H₂O: 1 mole
H₂SO₄: 1 mole

Being the molar mass of each compound:

SO₃: 80 g/mole
H₂O: 18 g/mole
H₂SO₄: 98 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

SO₃: 1 mole* 80 g/mole= 80 grams
H₂O: 1 mole* 18 g/mole= 18 grams
H₂SO₄: 1 mole* 98 g/mole= 98 grams

Then you can apply the following rule of three: if 1 mole of sulfuric acid is produced by the reaction of 80 grams of sulfur trioxide, 4 moles of sulfuric acid is produced from how much mass of sulfur trioxide?

$mass of sulfur trioxide= \frac{4 moles of sulfuric acid* 80 grams of sulfur trioxide}{1 mole of sulfuric acid }$

mass of sulfur trioxide= 320 grams

<u><em>320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.</em></u>

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2 years ago

Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

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<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

<u>Initial:</u> 0.1 0.1

<u>At eqllm:</u> 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

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3 years ago

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2 years ago

The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo

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Answer:

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

$V_1 = \frac{P_2V_2}{P_1} \\$

We have

$V_1 = \frac{7.98 \times 14.2}{3.14} = \frac{113.316}{3.14} \\ = 36.0878...$

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2 years ago

Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife

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Answer:

$\boxed{\text{2408 min}}$

Explanation:

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2. Calculate the half-life

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