How many atoms are in 13.97 liters of water vapor at STP

1 answer:

5 0

Let's assume that water vapor has ideal gas behavior. 
Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
T = 0 °C = 273 K (standard temperature)

By substitution,
101325 Pa x 13.97x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K
n = 0.624 mol

Hence, the moles of water vapor at STP is 0.624 mol.

According to the Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.

Hence, number of atoms in water vapor = 0.624 mol x 6.022 × 10²³ mol⁻¹
 = 3.758 x 10²³

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If you have 2.4L of SO2 gas (at STP) how many moles of sulfur dioxide do you have?

Andru [333]

Answer:

0.107 mole of SO2.

Explanation:

1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).

With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.

This can be obtained as follow:

22.4 L contains 1 mole of SO2.

Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.

Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.

4 0

3 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$