Question

In the reaction below, if a total of 10.0 g of zinc and hydrochloric acid react completely, what is the total mass of zinc chloride and hydrogen produced?Zn(s) + 2HCl(aq) to ZnCl2(aq) + H2(g)Exactly 10.0 gLess than 10.0 gNot enough information to tellGreater than 10.0 g

Answer 1

Answer:

Not enough information to tell.

Explanation:

What is given?

Mass of Zn = 10.0 g,

Mass of HCl = 10.0 g,

Molar mass of Zn = 65.4 g/mol,

Molar mass of HCl = 36.4 g/mol,

Molar mass of ZnCl2 = 136.2 g/mol,

Molar mass of H2 = 2 g/mol.

Step-by-step solution:

First, let's convert 10.0 g of each reactant to moles using their respective molar mass:

$\begin{gathered} 10.0\text{ g Zn}\cdot\frac{1\text{ mol Zn}}{65.4\text{ g Zn}}=0.153\text{ moles Zn,} \\ \\ 10.0\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=0.275\text{ moles HCl.} \end{gathered}$

Now, let's identify what is the limiting reactant. Let's see how many moles of ZnCl2 can be produced by 0.153 moles of Zn if 1 mol of Zn reacted produces 1 mol of ZnCl2, and how many moles of ZnCl2 can be produced by 0.275 moles of HCl if 2 moles of HCl reacted produces 1 mol of ZnCl2:

$\begin{gathered} 0.153\text{ moles Zn}\cdot\frac{1\text{ mol ZnCl}_2}{1\text{ mol Zn}}=0.153\text{ moles ZnCl}_2, \\ \\ 0.275\text{ moles HCl}\cdot\frac{1\text{ mol ZnCl}_2}{2\text{ moles HCl}}=0.138\text{ moles ZnCl}_2. \end{gathered}$

You can realize that the limiting reactant, in this case, is HCl because is the first reactant consumed first and this reactant 'impose' the limit to produce the products.

So now, let's find how many moles of H2 are being produced by 0.275 moles of HCl if 2 moles of HCl reacted produces 1 mol of H2:

$0.275\text{ moles HCl}\cdot\frac{1\text{ mol H}_2}{2\text{ moles HCl}}=0.138\text{ moles H}_2.$

The final step is to convert each number of moles of each product to grams using their respective molar mass, as follows:

$\begin{gathered} 0.138\text{ moles ZnCl}_2\cdot\frac{136.2\text{ g ZnCl}_2}{1\text{ mol ZnCl}_2}=18.8\text{ g ZnCl}_2, \\ \\ 0.138\text{ moles H}_2\cdot\frac{2\text{ g H}_2}{1\text{ mol H}_2}=0.276\text{ g H}_2. \end{gathered}$

We're producing 18.8 g of zinc chloride (ZnCl2), and 0.276 g of hydrogen (H2), so based on this logic the answer would be not enough information to tell.