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How many grams of zinc metal will react completely with 7.8 liters of 1.6 M HCl? Show all of the work needed to solve this problem

Zn (s) + 2HCl (aq)yields ZnCl2 (aq) + H2 (g)



How many grams of iron metal do you expect to be produced when 298 grams of an 83.1 percent by mass iron (II) nitrate solution react with excess aluminum metal? Show all of the work needed to solve this problem.
2Al (s) + 3Fe(NO3)2 (aq)yields 3Fe (s) + 2Al(NO3)3 (aq)

1 Answer

5 votes

Answer:

For 1: The correct answer is 407.97 grams.

For 2: The correct answer is 76.72 grams.

Step-by-step explanation:

  • For 1:

We are given the molarity of HCl, to find the moles of HCl, we use the formula:


Molarity=\frac{\text{Moles}{\text{Volume}}

We are given:

Molarity = 1.6 M

Volume = 7.8 L

Putting values in above equation, we get:


1.6mol/L=\frac{\text{Moles of HCl}}{7.8L}\\\\\text{Moles of HCl}=12.48mol

For the given reaction:


Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(aq.)+H_2(g)

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of zinc metal.

So, 12.48 moles of HCl react with =
(1)/(2)* 12.48=6.24mol of Zinc metal.

To calculate the mass of zinc metal, we use the formula:


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} .....(1)

Molar mass of Zinc metal = 65.38 g/mol


6.24=\frac{\text{Mass of zinc metal}}{65.38g/mol}\\\\\text{Mass of zinc metal}=407.97g

  • For 2:

We are given that 298 grams of 83.1 % by mass of iron (II) nitrate solution are present.

So, mass of Iron (II) nitrate solution will be =
(83.1)/(100)* 298=247.638g

Molar mass Iron (II) nitrate = 180 g/mol

Putting values in equation 1, we get:


\text{Moles of }Fe(NO_3)_2=(247.638g)/(180g/mol)=1.37mol

For the following reaction:


2Al(s)+3Fe(NO_3)_2(aq.)\rightarrow 3Fe(s)+2Al(NO_3)_3(aq.)

By Stoichiometry of the reaction:

3 moles of
Fe(NO_3)_2 produces 3 moles of iron metal.

So, 1.37 moles of
Fe(NO_3)_2 will produce =
(3)/(3)* 1.37=1.37mol of iron metal.

To calculate the mass of zinc metal, we equation 1:


1.37=\frac{\text{Mass of iron metal}}{56g/mol}\\\\\text{Mass of iron metal}=76.72g

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