2g of magnesium Oreo to excess dilute acid according to the equation Mg(s) + 2H (aq) --- Mg² (aq) + H2 (g) (Realitive atomic mass of Mg= 24, molar mass volume at room temperature and pressure = 24dm³) The volume of hydrogen, measured at room temperature I pressure, that is produced is A) 1000 cm³ B)1500 cm³ C) 2000 ³cm D) 24 000 cm³

Option C. 2000 cm³.

Explanation:

We'll begin by calculating the number of mole in 2 g of Mg. This is illustrated below:

Molar mass of Mg = 24 g/mol

Mass of Mg = 2 g

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 2/24

Mole of Mg = 8.33×10¯² mole.

Next, we shall determine the number of mole of H2 produced from the reaction. This is illustrated below:

Mg(s) + 2H^+ (aq) —› Mg^2+(aq) + H2 (g)

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Therefore, 8.33×10¯² mole of Mg will also produce 8.33×10¯² mole of H2.

Finally, we shall determine the volume of H2 produced. This is illustrated below:

1 mole of H2 occupy 24000 cm³ at room temperature and pressure.

Therefore, 8.33×10¯² mole of H2 will occupy = 8.33×10¯² x 24000 ≈ 2000 cm³ at room temperature and pressure.

Therefore, 2000 cm³ of H2 were obtained from the reaction.

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

the answer is d.) potassium