How many mL of 3.84 M HCL are required to consume 4.12g of Zinc in the following reaction?
Zn(s) + 2HCL(aq)----> ZnCl2(aq) + H2(g)
A method to work all stoichiometry problems.
1. Write the equation and balance it. Your's looks ok.
2. Convert what you are given into mols. In this case, mols Zn = grams/atomic mass.
3. Using the coefficients in the balanced equation, convert mols Zn to mols HCl.
4. Now that you have mols HCl, remember mols = M x L. You have mols and M, calculate L and convert to mL.
Post your work if you get stuck.
Your Steps made everything clear. Thank you so much :D
Well, I've got a zincing feeling that this is going to be a fun problem! Let's grab our calculators and get started.
First, we need to use the molar mass of zinc (Zn) to find the number of moles in 4.12 grams. The molar mass of zinc is around 65.38 g/mol. So, dividing 4.12 g by 65.38 g/mol, we get approximately 0.063 moles of zinc.
According to the balanced equation, for every 1 mole of Zn, we need 2 moles of HCl. So, we'll need 2 times 0.063 moles of HCl, which comes out to be around 0.126 moles.
Now comes the grand finale! To determine the volume, we can use the formula:
Volume (in liters) = Moles / Concentration
Converting 0.126 moles to liters, we divide it by the molar concentration of HCl, which is 3.84 M.
Volume (in liters) = 0.126 moles / 3.84 moles per liter
So, finally, the volume of 3.84 M HCl you'll need is approximately 0.03281 liters.
Now, if we convert liters to milliliters (1 liter = 1000 milliliters), the final answer is approximately 32.81 mL.
So, approximately 32.81 mL of 3.84 M HCl are required to consume 4.12g of zinc. Hope this helps tickle your funny bone while calculating!
To calculate the mL of 3.84 M HCl required to consume 4.12 g of zinc, you can use the stoichiometry of the reaction. Here's how you can do it step by step:
Step 1: Determine the molar mass of zinc (Zn).
The molar mass of zinc (Zn) can be found on the periodic table. It is approximately 65.38 g/mol.
Step 2: Convert the mass of zinc to moles.
To convert the mass of zinc (4.12 g) to moles, we divide it by the molar mass of zinc (65.38 g/mol).
Number of moles of Zn = 4.12 g / 65.38 g/mol = 0.063 moles
Step 3: Use stoichiometry to find the moles of HCl needed.
From the balanced equation, we can see that 1 mole of zinc reacts with 2 moles of HCl.
So, the moles of HCl required would be twice the moles of zinc.
Number of moles of HCl = 2 * 0.063 moles = 0.126 moles
Step 4: Calculate the volume of HCl solution needed.
Now that we have the moles of HCl required, we can use the molarity (3.84 M) to find the volume of HCl solution needed.
Molarity (M) = Moles of solute / Volume of solution (in liters)
Volume of solution (in liters) = Moles of solute / Molarity (M)
Volume of HCl solution = 0.126 moles / 3.84 mol/L = 0.0328 L
However, the volume is required in mL, so we need to convert it.
1 L = 1000 mL
Volume of HCl solution = 0.0328 L * 1000 mL/L = 32.8 mL
Therefore, approximately 32.8 mL of 3.84 M HCl is required to consume 4.12 g of zinc.
To determine the volume of HCl required to consume a given amount of Zinc, you need to use stoichiometry and the balanced equation for the reaction.
First, let's start by calculating the moles of Zinc (Zn) using its molar mass:
Molar mass of Zn = 65.38 g/mol
Given mass of Zn = 4.12 g
moles of Zn = given mass of Zn / molar mass of Zn
moles of Zn = 4.12 g / 65.38 g/mol
Next, we need to use the stoichiometry of the balanced equation to find the moles of HCl required to react with the calculated moles of Zinc.
From the balanced equation: Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)
we can see that 1 mole of Zinc (Zn) reacts with 2 moles of hydrochloric acid (HCl).
moles of HCl = 2 * moles of Zn
Now that we have the moles of HCl required, we can calculate the volume of the 3.84 M HCl solution needed. The concentration given is 3.84 moles/L (Molarity).
Here's the equation we can use:
moles of HCl = Molarity × volume (in liters)
Rearranging the equation to solve for the volume (in liters):
volume (in liters) = moles of HCl / Molarity
Let's calculate the volume:
volume (in liters) = moles of HCl / Molarity
volume (in liters) = (2 * moles of Zn) / 3.84 M
Finally, convert the volume from liters to milliliters (mL):
volume (in mL) = volume (in liters) x 1000
Now, you have all the information needed to calculate the volume of 3.84 M HCl required to consume 4.12 g of Zinc.