How many grams of zinc metal will react completely with 8.2 liters of 3.5 M HCl? Show all of the work needed to solve this problem.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
Here is a worked example of a stoichiometry problem. Just follow th steps. Remember moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To solve this problem, we need to use the concept of stoichiometry. Stoichiometry is a calculation method that allows us to determine the quantities of reactants and products in a chemical reaction.
First, we'll write down the balanced equation for the reaction between zinc and hydrochloric acid:
Zn (s) + 2 HCl (aq) -> ZnCl2 (aq) + H2 (g)
From the equation, we can see that one mole of zinc reacts with two moles of hydrochloric acid to produce one mole of zinc chloride and one mole of hydrogen gas.
Next, we need to calculate the number of moles of hydrochloric acid (HCl) present in the given solution. We have the concentration of HCl, which is 3.5 M, and the volume of the solution, which is 8.2 liters.
To calculate the number of moles, we use the formula:
moles = concentration × volume
moles of HCl = 3.5 M × 8.2 L
moles of HCl = 28.7 mol
Since the reaction between Zn and HCl occurs in a 1:2 ratio, one mole of zinc reacts with two moles of hydrochloric acid.
To determine the number of moles of zinc required, we divide the moles of HCl by 2:
moles of Zn = 28.7 mol ÷ 2
moles of Zn = 14.35 mol
Now that we have the number of moles of zinc required, we can use its molar mass to calculate the mass of zinc. The molar mass of zinc is 65.38 g/mol.
mass of Zn = moles of Zn × molar mass of Zn
mass of Zn = 14.35 mol × 65.38 g/mol
mass of Zn = 938.623 g
Therefore, approximately 938.623 grams of zinc metal will react completely with 8.2 liters of 3.5 M hydrochloric acid.