How many milliliters of 4.00 M HCl(aq) are required to react with 5.25 g Zn(s)?

Answer 1

The first step is to write out the reaction:

HCl (aq) + Zn (s) -> ZnCl₂ + H₂

Notice we still have to balance the equation...

The reaction is now the following:

2HCl (aq) + Zn (s) -> ZnCl₂ + H₂

Now we can figure out how to determine the milliliters of HCl needed using dimensional analysis. Because our endpoint is for HCl, we start with 5.25 g of Zn.

5.25 g of Zn can be converted to mol of Zn using its molar mass from the periodic table, and it is 65.39 g/mol.

5.25 g Zn/65.39 g/mol = 0.0803 mol Zn

In order to find mol of HCl, we must use the equation to convert from mol of Zn to mol of HCl. Notice that there is a 2:1 ratio for mol of HCl to mol of Zn.

Therefore, there is 2 x 0.0803 mol = 0.1606 mol HCl.

Now that we have mol of HCl, we can use the given molarity (4.00 M or 4.00 mol/L) to calculate the milliliters of HCl.

0.1606 mol / 4.00 mol/L = 0.04015 L

We currently have our answer in liters (L), but we need to convert to milliliters by multiplying by 1000 or moving the decimal place 3 times.

The final answer is 40.15 milliliters (mL), but in order to have 3 significant figures, it would be 40.2 mL if we round up.

Let me know if you have any additional questions!