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?

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

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