how many milliliters of a 0.46 M HCl solution are needed to react completely with a 6.4 g of zinc to form zinc (ii) chloride answer in mL
Help please :( I was absent and don't know where to start
Why did the acid go to therapy? Because it had a pH imbalance! Okay, let's get started on your question. First, we need to calculate the number of moles of zinc using its molar mass. Zinc (Zn) has a molar mass of 65.38 g/mol. So, 6.4 g of zinc is equal to 6.4 g / 65.38 g/mol = 0.0979 mol.
The balanced equation for the reaction between HCl and zinc is:
Zn + 2HCl -> ZnCl2 + H2.
The stoichiometry of the balanced equation tells us that 1 mol of zinc reacts with 2 mol of HCl, so we need twice as many moles of HCl as zinc.
Therefore, we need 2 * 0.0979 mol = 0.1958 mol of HCl.
Now, let's use the given concentration of the HCl solution (0.46 M) to determine the volume of solution needed. The equation for molarity is:
Molarity (M) = moles of solute / volume of solution (L).
We want to find the volume in milliliters (mL), so we need to rearrange the equation:
Volume of solution (L) = moles of solute / Molarity (M).
Plugging in the values, we have:
Volume of solution (L) = 0.1958 mol / 0.46 M = 0.4256 L.
Lastly, let's convert liters to milliliters:
Volume of solution (mL) = 0.4256 L * 1000 mL/L = 425.6 mL.
So, you would need approximately 425.6 mL (or round it to 426 mL) of the 0.46 M HCl solution to react completely with 6.4 g of zinc.
To solve this problem, you will need to use stoichiometry and the balanced chemical equation for the reaction between zinc and hydrochloric acid.
First, let's write the balanced chemical equation for the reaction:
Zn + 2HCl → ZnCl2 + H2
According to the balanced equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of ZnCl2.
Step 1: Convert the given mass of zinc (6.4 g) to moles using the molar mass of zinc (65.38 g/mol).
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 6.4 g / 65.38 g/mol
Step 2: Once you have the moles of Zn, use the mole ratio from the balanced equation to determine the moles of HCl required.
moles of HCl = moles of Zn * (2 moles of HCl / 1 mole of Zn)
Step 3: Finally, calculate the volume of the HCl solution using the molarity of the HCl solution (0.46 M) and the moles of HCl.
volume of HCl solution (in liters) = moles of HCl / molarity of HCl
volume of HCl solution (in milliliters) = volume of HCl solution (in liters) * 1000 mL/L
Now, let's plug in the values and solve the problem:
Step 1:
moles of Zn = 6.4 g / 65.38 g/mol
moles of Zn = 0.09794 mol
Step 2:
moles of HCl = 0.09794 mol * (2 mol HCl / 1 mol Zn)
moles of HCl = 0.19588 mol
Step 3:
volume of HCl solution (in liters) = 0.19588 mol / 0.46 M
volume of HCl solution (in liters) = 0.426 mL
Therefore, you would need approximately 0.426 mL of the 0.46 M HCl solution to react completely with 6.4 g of zinc to form zinc (II) chloride.
To solve this problem, you need to use the concept of stoichiometry and the molar ratio between the reactants, HCl and zinc.
First, let's calculate the moles of zinc using its molar mass. The molar mass of zinc is 65.38 g/mol.
So, moles of zinc = mass of zinc / molar mass of zinc
= 6.4 g / 65.38 g/mol
= 0.098 mol
Now, let's establish the balanced chemical equation for the reaction between HCl and zinc:
2 HCl + Zn -> ZnCl2 + H2
From the balanced equation, we can see that every 2 moles of HCl react with 1 mole of zinc to produce 1 mole of zinc chloride (ZnCl2).
Therefore, the moles of HCl needed to react completely with the given amount of zinc will also be 0.098 mol.
To find the volume of the 0.46 M HCl solution needed, we can use the equation:
moles of solute = molarity × volume of solution (in liters)
Rearranging the equation, we get:
volume of solution (in liters) = moles of solute / molarity
Substituting the values, we have:
volume of solution (in liters) = 0.098 mol / 0.46 mol/L
= 0.213 L
Finally, we can convert the volume from liters to milliliters by multiplying by 1000:
volume of solution (in milliliters) = 0.213 L × 1000 mL/L
= 213 mL
Therefore, you will need 213 mL of the 0.46 M HCl solution to completely react with 6.4 g of zinc to form zinc (II) chloride.
start by reading your text book. I'm sure there's a section regarding moles.
Zn has a mol wt of 65.39, so 6.4g = 0.1 moles of Zn
Since the reaction is
Zn + 2HCl = ZnCl2 + H2
each mole of Zn reacts with 2 moles HCl
So, you need .2 moles HCl
.46M HCl has .46 moles/liter
So, you need .2/.46 = .434 liters, or 434ml