According to the law of conservation of energy,

A compound is analyzed and found to be 50% sulfur and 50% oxygen. If the total amount of the sulfur oxide is 12.5 g, how many gr

il63 [147K]

Answer:

25.0g is the mass of sulfur

Explanation:

The sulfur and the oxygen are 50:50. That means there is 1 mole of S per mole of O.

To solve this question we need to convert the mass of O to moles, as moles of O = Moles of S, we can find the moles of S and its mass:

Moles O = Moles S:

12.5g O * (1mol / 16g) = 0.781 moles

Mass S:

0.781 moles * (32g / mol) =

<h3>25.0g is the mass of sulfur</h3>

7 0

3 years ago

What does Einstein's famous equation for nuclear energy, E = mc2, mean?

xz_007 [3.2K]

Some mass is converted into energy in a nuclear reaction. Hope this helps!

6 0

3 years ago

Read 2 more answers

Which box depicts the tissue level of organization? A) P B) Q C) R D) S

Nuetrik [128]

Which box depicts the tissue level of organization? A) P B) Q C) R D) S

The answer is B) Q

4 0

4 years ago

What are the steps needed when solving problem like this?

emmainna [20.7K]

For the answer to the question above,the gas can be decomposed to yield 2.09 parts by mass of uranium for every 1 part by mass of fluorine.If the relative mass of a uranium atom is 238 and the relative mass of a fluorine atom is 19, the number of fluorine atoms that are combined with one uranium atom. is 6

3 0

3 years ago

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Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: CO(g)+2H2(g)→CH3OH(g)CO(g)+2H2(g)→CH3O

zvonat [6]

Answer: The limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$ ..........(1)

For carbon monoxide:

We are given:

$P=232mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$232mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{232\times 1.55}{62.3637\times 305}=0.0189mol$

For hydrogen gas:

We are given:

$P=389mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$389mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{389\times 1.55}{62.3637\times 305}=0.0317mol$

For the given chemical equation:

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of carbon monoxide

So, 0.0317 moles of hydrogen gas will react with = $\frac{1}{2}\times 0.0317=0.01585mol$ of carbon monoxide

As, given amount of carbon monoxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 1 mole of methanol

So, 0.0317 moles of hydrogen gas will produce = $\frac{1}{2}\times 0.0317=0.01585moles$ of methanol

Now, calculating the mass of methanol from equation 1, we get:

Molar mass of methanol = 32 g/mol

Moles of methanol = 0.01585 moles

Putting values in equation 1, we get:

$0.01585mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.01585mol\times 32g/mol)=0.507g$

Hence, the limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams

7 0

3 years ago