g Potassium metal reacts with water to give potassium hydroxide and hydrogen gas. If 75.5 mL of hydrogen gas is produced at STP, what is the mass of potassium metal that reacted

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.

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3