Question

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)

Answer 1

Answer:

For 1: The correct answer is 407.97 grams.

For 2: The correct answer is 76.72 grams.

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 = $\frac{1}{2}\times 12.48=6.24mol$ 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 = $\frac{83.1}{100}\times 298=247.638g$

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

Putting values in equation 1, we get:

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

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 = $\frac{3}{3}\times 1.37=1.37mol$ of iron metal.

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

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