Zinc metal reacts with hydrochloric acid, according to the chemical equation below.

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Tina added 0.146 mole of hydrochloric acid solution to 0.135 mole of zinc in a beaker and waited several minutes for the reaction to complete. Which of the following best describes the reaction that Tina performed?
A
HCl was the limiting reagent, and 0.0675 mole of ZnCl2 was produced.

B
HCl was the limiting reagent, and 0.0730 mole of ZnCl2 was produced.

C
Zn was the limiting reagent, and 0.135 mole of ZnCl2 was produced.

D
Zn was the limiting reagent, and 0.0730 mole of ZnCl2 was produced.

Option B. HCl was the limiting reagent, and 0.0730 mole of ZnCl2 was produced.

Explanation:

We'll begin by obtaining writing the equation for the reaction. This is shown below:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Data obtained from the question:

Mole of HCl = 0.146 mole

Mole of Zn = 0.135 mole

Now let us determine which is the limiting reactant. This is illustrated below below:

From the equation above,

2 mole of HCl required 1 mole of Zn.

Therefore, 0.146 mole of HCl will require = 0.146/2 = 0.073 mole of Zn.

Now this amount ( i.e 0.073 mole) of Zn obtained is far lower than the amount (i.e 0.135 mole) of Zn that was given. From this calculation, HCl is the limiting reactant as all the amount of HCl was used up and Zn is the excess reactant.

Now let us calculate the amount of the product formed from 0.146 mole of HCl. This is illustrated below:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

From the equation above,

2 moles of HCl produced 1 mole of ZnCl2.

Therefore, 0.146 mole of HCl will produce = 0.146/2 = 0.073 mole of ZnCl2.

Therefore, HCl is the limiting reactant and 0.073 mole of ZnCl2 was produced.

Option B is correct

HCl was the limiting reagent, and 0.0730 mole of ZnCl2 was produced.

Explanation:

Step 1: data given

Number of moles hydrochloric acid (HCl) = 0.146 moles

Number of moles of zinc (Zn) = 0.135 moles

Molar mass HCl = 36.46 g/mol

Atomic mass Zn = 65.38 g/mol

Step 2: The balanced equation

2HCl + Zn → ZnCl2 + H2

Step 3: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Zn to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will all be consumed (0.146 moles). Zn is in excess. There will react 0.146/2 = 0.073 moles. There will remain 0.135-0.072 = 0.062

Step 4: Calculate moles of products

For 2 moles HCl we need 1 mol Zn to produce 1 mol ZnCl2 and 1 mol H2

For 0.146 moles HCl we'll have 0.146 / 2  = 0.073 moles ZnCl2 and 0.146/2 = 0.073 moles H2

Option B is correct

HCl was the limiting reagent, and 0.0730 mole of ZnCl2 was produced.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa