Question

Balance the following Equation:

___Zn + ___HCl --> ___ZnCl2 + ___H2
50g of zinc are reacted with 50g of hydrogen chloride. Calculate the amount of excess reactant.


A- 4.2g

B- 5.1g

C- 1.4g

D- 2.3g

Answer 1

Answer:

The amount of excess is 5.16 grams. Option B 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 n 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 remain 0.764 -0.685 = 0.079 moles

Step 5: Calculate the remaining mass of zinc

Mass = moles * molar mass

Mass Zinc = 0.079 moles *65.38 g/mol

Mass Zinc = 5.16 grams

The amount of excess is 5.16 grams. Option B is correct

Answer 2

Following are the calculating steps for the given equation:

For step 1:

Given Data:

Mass of $(Zn) = 50.0\ grams$

Mass of $(HCl) = 50.0 \ grams$

Calculating the atomic mass $Zn = 65.38\ \frac{g}{mol}$

Calculating the Molar mass $HCl = 36.46 \ \frac{g}{mol}$

For step 2:

Calculating the equation balancing:

$\bold{Zn + 2HCl \to ZnCl_2 + H_2}$

For step 3:

Calculating the moles:

$\to Moles = \frac{mass}{molar\ mass}$

Moles of $Zn= \frac{50.0\ grams}{65.38\ \frac{g}{mol}} = 0.764\ moles$

Moles of $HCl = \frac{50.0\ grams}{ 36.46\ \frac{g}{mol}}= 1.37\ moles$

For step 4:

Calculating the limiting of the reactant:

• When the $1 \ mol\ Zn$ we need $2\ moles\ HCl$ to produce ${ 1\ mol\ ZnCl_2 \ and\ 1 \ mol \ H_2}$ $HCl$limiting reactant.
• It will absolutely consume ($1.37\ moles$) $Zn$ is in excess.
• Calculating the react value $\ \frac{1.37}{2} = 0.685\ moles$

            $\to 0.764 -0.685 = 0.079\ moles$

For step 5:

Calculating the mass of $Zn$:

Formula:

$\to \text {Mass of (Zn) = moles} \times \text{molar mass}$

                      $= 0.079\ moles \times 65.38\ \frac{g}{mol}\\\\= 5.16\ grams$

Calculating the excess amount of $\ \bold{5.16}\ grams$.

Therefore, the final answer is "Option B".

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