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BigorU [14]
3 years ago
8

How many grams of sulfur trioxide are produced 18 mol O2 react with sufficient sulfur? Show all your work S8 +12O 2> 8SO3

Chemistry
1 answer:
SOVA2 [1]3 years ago
7 0

Answer:

m_{SO_3}=2.31x10^4gSO_3

Explanation:

Hello!

In this case, since the reaction between sulfur and oxygen is:

S_8 +\frac{1}{2} O_2 \rightarrow 8SO_3

Whereas there is a 1/2:8 mole ratio between oxygen and SO3, and we can compute the produced grams of product as shown below:

m_{SO_3}=18molO_2*\frac{8molSO_3}{1/2molO_2} *\frac{80.06gSO_3}{1molSO_3} \\\\m_{SO_3}=23,057.3gSO_3=2.31x10^4gSO_3

Best regards!

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Read 2 more answers
Need help !!!!! ASAP
Oksi-84 [34.3K]
<h2>Hello!</h2>

The answer is:

The approximate volume is 9.84 L.

V=9.84L

<h2>Why?</h2>

Since we are given the numer of moles, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

PV=nRT

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant, which is equal to:

R=0.082\frac{atm.L}{mol.K}

So, we are given the information:

n=1.5mol\\Temperature=300K\\Pressure=3.75atm

Now, substituting the given information and isolating the Volume from The Ideal Gas Law equation, we have:

PV=nRT

V=\frac{nRT}{P}

3.75atm*V=1.5mol*0.082*\frac{atm.L}{mol.K}*300K

V=\frac{ 1.5mol*0.082*\frac{atm.L}{mol.K}*300K}{3.75atm}\\\\V=\frac{36.9atm.L}{3.75atm}=9.84L

So, the approximate volume is 9.84 L.

V=9.84L

Have a nice day!

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4 years ago
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