Give the equation: 2k + 2h2o --> 2koh + h2 if 23.5 grams of potassium are reacted with excess water, how many grams of hydrogen will be formed?

Answers

Answers:

0.606 g H₂

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:   39.10                            2.016

           2K + 2H₂O ⟶ 2KOH + H₂

m/g:   23.5

(a) Calculate the moles of K

n = 23.5 g K × 1 mol K /39.10 g K

n = 0.6010 mol K

(b) Calculate the moles of H₂

The molar ratio is (1 mol H₂/2 mol K)

n = 0.6010 mol K × (1 mol H₂/2 mol K)

n = 0.3005 mol H₂

(c) Calculate the mass of H₂

m = 0.3005 mol H₂ × (2.016 g H₂/1 mol H₂)

m = 0.606 g H₂

0.6058g of Hydrogen wouldbe produced

Explanation:

2K + 2H2O --> 2KOH + H2

Our focus is on potassium and Hydrogen.

The question stated that water in excess, this means that water is not a limiting reactant and what that means is that the amount of product formed is dependent on potassium and not water.

From the equation,

2 mol of potassium produces 1 mol of Hydrogen;

Expressing in terms of mass;

Note that: Mass = Number of moles * Molar mass

78.1966g ( 2 * 39.0983) of potassium produces 2.01588g ( 1 * 2.01588) of hydrogen Hydrogen

How many grams of Hydrogen would then be produced from 23.5g of potassium?

78.1966 = 2.01588

23.5 = x

Upon cross multiplication and solving for x;

x = ( 23.5 * 2.01588) / 78.1966

x = 0.6058g

.605862g H2

To find this, set up the following equation:

You're starting with 23.5 grams of potassium and you want to end up with grams of Hydrogen. To do that, you need to do a molar ratio:

23.5gK \frac{1 Mol K}{39.0983g K}\frac{1 Mol H2}{2 Mol K}\frac{2.016g H2}{1 Mol H2}

Multiply everything on the top (23.5 x 1 x 1 x 2.016) and on the bottom (39.098 x 2 x 1).

Divide the product given from the top, 47.376, by the product from the bottom, 78.196 to get the final mass of .605862 or rounded up to .606

The reason for the molar ratio is because in the given equation, you need to take into account the needed amount of moles for each element/compound. You also cannot just go from grams to grams, you must have a conversion of moles in between just as you would for grams to molecules.

Hope this helps!



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