What is the volume of 2 mol of chlorine gas at STP? 2.0 L 11.2 L 22.4 L 44.8 L

Chemistry

1 answer:

nirvana33 [79]3 years ago

7 0

Answer:

44.8 L

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At Standard temperature and pressure (STP);

P = 1 atm

T = 273K

Hence, when n = 2moles, the volume of the gas is:

Using PV = nRT

1 × V = 2 × 0.0821 × 273

V = 44.83

V = 44.8 L

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Read 2 more answers

Given the following equation: 2K + Cl2 -> 2KCl How many grams of KCl is produced from 4.00 g of K and excess Cl2?

Thepotemich [5.8K]

Answer:

42.65g

Explanation:

Given parameters:

Mass of K = 4g

Unknown: Mass of KCl

Solution:

Complete equation of the reaction:

2K + Cl₂ → 2KCl

To solve this problem, we know that the reactant in short supply is potassium K and this dictates the amount of products that would be formed. The chlorine gas is in excess and we can't use it to determine the amount of product that would form.

Now, we work from the known to the unknown. Since we know the mass of K given in the reaction, we can simply find the molar relationship between the reacting potassium and the product. We simply convert the mass to mole and compare to the product. From there we can find the mass of KCl that would be produced.

Calculating number of moles of K

Number of moles = $\frac{mass}{molar mass}$