How many milliliters of 2.00 M HCl(aq) are required to react with 7.55 g of an ore containing 36.0% Zn(s) by mass?
1. construct a balance equation for HCl and Zn reaction and note the mole ratio between HCl and Zn.
2. calculate the mass of Zn from the ore i.e. 36.0% of 7.55g gives the mass of Zinc.
3. Calculate the mole of zinc using n = m/Mr where Mr is the molar mass of zinc and m is the mass in grams from 2.
4. Use the mole of zinc in 3 to determine the mole of HCl using the mole ratio in 1.
5. with the mole in 4, calculate the v using the formula M = n/v where M is the molarity of HCl, n is the mole in 4 and v is the volume in liters (you need to convert to mL).
hope that helps.
To calculate the volume of HCl(aq) required to react with the Zn(s) in the ore, we need to follow these steps:
Step 1: Determine the molar mass of Zn(s)
Zinc (Zn) has a molar mass of 65.38 g/mol.
Step 2: Calculate the mass of Zn(s) in the ore
To do this, multiply the mass of the ore (7.55 g) by the mass percent of Zn(s) in the ore (36.0%):
Mass of Zn(s) = 7.55 g × 0.36 = 2.718 g
Step 3: Convert the mass of Zn(s) to moles
To convert grams to moles, divide the mass (in grams) by the molar mass:
Moles of Zn(s) = 2.718 g / 65.38 g/mol ≈ 0.04159 mol
Step 4: Determine the stoichiometric ratio between Zn(s) and HCl(aq)
From the balanced chemical equation, we can see that the ratio between Zn(s) and HCl(aq) is 1:2. This means that 1 mole of Zn(s) reacts with 2 moles of HCl(aq).
Step 5: Calculate the moles of HCl(aq) needed
Multiply the moles of Zn(s) calculated in step 3 by the stoichiometric ratio:
Moles of HCl(aq) = 0.04159 mol Zn(s) × (2 mol HCl(aq) / 1 mol Zn(s)) ≈ 0.08318 mol
Step 6: Convert moles of HCl(aq) to milliliters
To convert moles to milliliters, we need to use the molarity (2.00 M) of the HCl(aq).
First, multiply the moles of HCl(aq) by the molarity to get the number of millimoles:
Millimoles of HCl(aq) = 0.08318 mol × (1000 mmol/mol) ≈ 83.18 mmol
Finally, divide the millimoles by the molarity to get the volume in milliliters:
Volume of HCl(aq) = 83.18 mmol / (2.00 mmol/mL) ≈ 41.59 mL
Therefore, approximately 41.59 milliliters of 2.00 M HCl(aq) are required to react with 7.55 g of the ore containing 36.0% Zn(s) by mass.
To answer this question, we need to follow a series of steps:
Step 1: Calculate the molar mass of Zn(s):
The molar mass of Zn is 65.38 g/mol (from the periodic table). Since the ore contains 36.0% Zn by mass, we can calculate the mass of Zn present in the ore:
Mass of Zn = (36.0/100) * 7.55 g = 2.718 g
Step 2: Calculate the number of moles of Zn:
By using the molar mass of Zn, we can calculate the number of moles:
Number of moles of Zn = Mass of Zn / Molar mass of Zn
= 2.718 g / 65.38 g/mol
= 0.0416 mol
Step 3: Determine the stoichiometry of the reaction:
From the balanced chemical equation, we can determine the stoichiometric ratio between Zn and HCl. The reaction equation is:
Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)
From the equation, we see that 1 mole of Zn reacts with 2 moles of HCl. Therefore, the stoichiometric ratio is:
1 mole Zn : 2 moles HCl
Step 4: Calculate the number of moles of HCl required:
Using the stoichiometric ratio, we can calculate the number of moles of HCl required to react:
Number of moles of HCl = 2 * Number of moles of Zn
= 2 * 0.0416 mol
= 0.0832 mol
Step 5: Calculate the volume of HCl required:
We know that the concentration of HCl is 2.00 M, which means that there are 2.00 moles of HCl in every liter of the solution. Therefore, to calculate the volume required, we can use the following formula:
Volume of HCl (in liters) = Number of moles of HCl / Concentration of HCl
= 0.0832 mol / 2.00 mol/L
= 0.0416 L
Step 6: Convert the volume from liters to milliliters:
Since the question asks for the answer in milliliters, we need to convert the volume from liters to milliliters by multiplying it by 1000:
Volume of HCl (in milliliters) = Volume of HCl (in liters) * 1000
= 0.0416 L * 1000
= 41.6 mL
Therefore, approximately 41.6 milliliters of 2.00 M HCl(aq) are required to react with 7.55 g of the ore containing 36.0% Zn(s) by mass.