Question

8.
Fe(s) + 2HC1 FeC1 + H
(aq)
2(aq)
2(g)
When a student adds 30.0 mL of 1.00 M HCl to 0.56 g of powdered Fe, a reaction occurs
according to the equation above.
When the reaction is complete at 273 K and 1.0 atm, which of the following is true?
A
B
C
D
HCl is in excess, and 0.100 mol of HCl remains unreacted.
HCl is in excess, and 0.020 mol of HCl remains unreacted.
0.015 mol of FeC12 has been produced.
0.22 L of H2 has been produced.

Answer 1

Answer:

B

Explanation:

To determine the correct answer, we need to first balance the chemical equation:

Fe(s) + 2HCl(aq) -> FeCl2(aq) + H2(g)

Based on the balanced equation, we can determine the stoichiometry of the reaction.

From the balanced equation, we can see that 1 mole of Fe reacts with 2 moles of HCl to produce 1 mole of FeCl2 and 1 mole of H2.

Given that there is 0.56 g of Fe, we can calculate the moles of Fe using its molar mass:

moles of Fe = mass of Fe / molar mass of Fe = 0.56 g / 55.85 g/mol ≈ 0.010 mol

From the stoichiometry of the balanced equation, we can see that 0.010 mol of Fe will react with 0.020 mol of HCl (2 moles of HCl for every 1 mole of Fe).

Since the student added 30.0 mL of 1.00 M HCl, we can calculate the moles of HCl added:

moles of HCl = volume of HCl (in L) * molarity of HCl = 0.030 L * 1.00 mol/L = 0.030 mol

Comparing the moles of HCl added (0.030 mol) with the stoichiometry of the balanced equation (0.020 mol), we can determine that HCl is in excess and 0.020 mol of HCl remains unreacted.

Therefore, the correct answer is:

B) HCl is in excess, and 0.020 mol of HCl remains unreacted.

mark me as brilliant