Question

1. A 2.25 g sample of scandium (Sc) metal is reacted with excess hydrochloric acid to produce 2.41 L of hydrogen gas at 100°C and 0.950 atm. Determine the formula of the ""scandium chloride"" produced in the reaction.

Answer 1

The formula of scandium chloride is ScCl3

Step-by-step explanation:

Step 1: Data given

Mass of scandium = 2.25 grams

Volume of hydrogen gas produce = 2.41 L

Temperature = 100 °C = 373 K

Pressure = 0.950 atm

Step 2: Calculatem oles of scandium

Moles Sc = mass Sc / molar mass Sc

Moles Sc = 2.25 grams / 44.955 g/mol

Moles Sc = 0.050 moles

Step 3: Calculate moles hydrogen gas

p*V = n*R*T

⇒with p = the pressure of H2 gas = 0.950 atm

⇒with V = the volume of H2 gas = 2.41 L

⇒with n = the number of moles H2 gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature 373 K

n = (p*V)/(R*T)

n = (0.950 * 2.41)/(0.08206*373)

n = 0.0748 moles

Step 4: Calculate mole ratio

n(Sc)/n(H2) = 0.05 moles / 0.0748 moles

n(Sc)/ n(H2) = 2/3

This means for 2 moles Sc we'll have 3 moles H2

2Sc + HCl → ScCl + 3H2

Since we have 6x H on the right side, we have to multiply HCl on the left side by 6. To balance the amount of H.

2Sc + 6HCl → ScCl + 3H2

Since we have 2x Sc on the left side, and 1x Sc (in ScCl) on the right (in ScCl) we have to multiply ScCl on the right side by 2

2Sc + 6HCl → 2ScCl + 3H2

This means We have 3x Cl in the scandium chloride (ScCl3)

The formula of scandium chloride is ScCl3