Respuesta :
Answer:
Zn is the excess reactant (0.079 moles in excess)
Option C is correct
Explanation:
Step 1: Data given
Mass of Zinc (Zn) = 50.0 grams
Mass of Hydrogen chloride (HCl) = 50.0 grams
atomic mass Zn = 65.38 g/mol
Molar mass HCl = 36.46 g/mol
Step 2: The balanced equation
Zn + 2HCl → ZnCl2 + H2
step 3: Calculate moles
Moles = mass / molar mass
Moles Zn = 50.0 grams / 65.38 g/mol
Moles Zn = 0.764 moles
Moles HCl = 50.0 grams / 36.46 g/mol
Moles HCl = 1.37 moles
Step 4: Calculate limiting reactant
For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2
HCl is the limiting reactant. It will completely be consumed (1.37 moles)
Zn is in excess. There will react 1.37/2 = 0.685 moles
There will be 0.764 -0.685 = 0.079 moles in excess
Following are the calculating steps for the given equation:
For step 1:
Given Data:
Mass of [tex](Zn) = 50.0\ grams[/tex]
Mass of [tex](HCl) = 50.0 \ grams[/tex]
Calculating the atomic mass [tex]Zn = 65.38\ \frac{g}{mol}[/tex]
Calculating the Molar mass [tex]HCl = 36.46 \ \frac{g}{mol}[/tex]
For step 2:
Calculating the equation balancing:
[tex]\bold{Zn + 2HCl \to ZnCl_2 + H_2}[/tex]
For step 3:
Calculating the moles:
[tex]\to Moles = \frac{mass}{molar\ mass}[/tex]
Moles of [tex]Zn= \frac{50.0\ grams}{65.38\ \frac{g}{mol}} = 0.764\ moles[/tex]
Moles of [tex]HCl = \frac{50.0\ grams}{ 36.46\ \frac{g}{mol}}= 1.37\ moles[/tex]
For step 4:
Calculating the limiting of the reactant:
- When the [tex]1 \ mol\ Zn[/tex] we need [tex]2\ moles\ HCl[/tex] to produce [tex]{ 1\ mol\ ZnCl_2 \ and\ 1 \ mol \ H_2}[/tex] [tex]HCl[/tex] is the limiting reactant.
- It will absolutely consume ([tex]1.37\ moles[/tex]) [tex]Zn[/tex] is in excess.
- Calculating the react value [tex]\ \frac{1.37}{2} = 0.685\ moles\\\\[/tex]
[tex]\to 0.764 -0.685 = 0.079\ moles[/tex]
Therefore, the final answer is "Option C".
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