Respuesta :
Answer:
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.
Answer:
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.