Answer:
0.26g of potassium, K.
Explanation:
We'll begin by calculating the number of mole of Hydrogen gas, H2 that will occupy 75.5mL at stp. This can be obtained as follow:
1 mole of a gas occupies 22.4L or 22400mL at stp. This implies that:
1 mole of H2 occupied 22400mL at stp.
Therefore, Xmol will occupy 75.5mL at stp i.e
Xmol of H2 = 75.5/22400
Xmol of H2 = 3.37×10¯³ mole
Therefore, 3.37×10¯³ mole of H2 occupied 75.5mL
Next, we shall determine the number of mole of potassium, K that is required to produce 3.37×10¯³ mole of H2. This is illustrated below:
2K + 2H2O —> 2KOH + H2
From the balanced equation above,
2 moles of K reacted to produce 1 mole of H2.
Therefore, Xmol of K will react to produce 3.37×10¯³ mole of H2 i.e
Xmol of K = 2 x 3.37×10¯³
Xmol of K = 6.74×10¯³ mole
Therefore, 6.74×10¯³ mole of K is needed to produce 3.37×10¯³ mole of H2.
Finally, we shall convert 6.74×10¯³ mole of K to grams. This can be achieved as shown below:
Molar mass of K = 39g/mol
Mole of K = 6.74×10¯³ mole
Mass of K =?
Mole = mass/molar mass
6.74×10¯³ = mass of K /39
Cross multiply
Mass of K = 6.74×10¯³ x 39
Mass of K = 0.26g
Therefore, 0.26g of K is needed for the reaction.
