determine the volume of 0.250 m H2SO4 needed to react with 5.00 g of zinc to produce H2 and ZnSO4
I'm sure you meant o.250 M. M stands for molarity while m stands for molality. There is a difference.
Zn + H2SO4 ==> ZnSO4 + H2
mols Zn = grams/atomic mass = 5.00/about 65 = about 0.076
From the coefficients you can tell it takes 1 mols H2SO4 for every 1 mole Zn used; therefore, you will need about 0.076 = about 0.076 mols H2SO4.
M H2SO4 = mols/L
0.250 M = 0.076/L. Solve for L and I would convert to mL.
Polst your work if you get stuck. Confirm the atomic mass of Zn and check all the calculations.
To determine the volume of 0.250 m H2SO4 needed to react with 5.00 g of zinc, we need to use the balanced chemical equation for the reaction between zinc and sulfuric acid.
The balanced equation is:
Zn + H2SO4 -> H2 + ZnSO4
From the equation, we can see that the ratio between zinc (Zn) and sulfuric acid (H2SO4) is 1:1.
First, we need to determine the number of moles of zinc. To do this, we'll use the molar mass of zinc (65.38 g/mol):
Number of moles of Zn = mass of Zn / molar mass of Zn
= 5.00 g / 65.38 g/mol
≈ 0.0764 mol
Since the ratio between Zn and H2SO4 is 1:1, we know that we need the same number of moles of sulfuric acid.
Next, we'll determine the volume of 0.250 m H2SO4 needed. The "molarity" (M) of a solution is defined as the number of moles of solute per liter of solution. So, 0.250 M H2SO4 means there are 0.250 moles of H2SO4 in 1 liter of solution.
Volume of H2SO4 = Number of moles of H2SO4 / Molarity of H2SO4
= 0.0764 mol / 0.250 mol/L
= 0.3056 L
≈ 305.6 mL
Therefore, approximately 305.6 mL of 0.250 m H2SO4 is needed to react with 5.00 g of zinc to produce H2 and ZnSO4.
To determine the volume of 0.250 M H2SO4 needed to react with 5.00 g of zinc, we need to use the balanced chemical equation for the reaction and perform stoichiometry calculations.
The balanced chemical equation for the reaction is:
Zn + H2SO4 -> H2 + ZnSO4
1 mole of zinc (Zn) reacts with 1 mole of sulfuric acid (H2SO4) to produce 1 mole of hydrogen gas (H2) and 1 mole of zinc sulfate (ZnSO4).
First, we need to determine the number of moles of zinc we have:
Using the atomic mass of zinc (Zn):
1 mole of Zn = 65.38 g (atomic mass of Zn)
Given that we have 5.00 g of zinc, we can calculate the number of moles:
Number of moles of Zn = Mass / Atomic mass = 5.00 g / 65.38 g/mol ≈ 0.076 moles
Since the stoichiometry of the reaction tells us that 1 mole of Zn reacts with 1 mole of H2SO4, we can conclude that we need 0.076 moles of H2SO4.
Next, we need to calculate the volume of 0.250 M H2SO4 required. The concentration of 0.250 M means that there are 0.250 moles of H2SO4 per liter of solution.
Number of moles of H2SO4 = volume in liters × concentration
0.076 moles = volume in liters × 0.250 mol/L
Solving for the volume in liters:
volume in liters = 0.076 moles / 0.250 mol/L ≈ 0.304 L
Finally, convert liters to milliliters:
volume in milliliters = 0.304 L × 1000 mL/L ≈ 304 mL
Therefore, approximately 304 mL of 0.250 M H2SO4 is needed to react with 5.00 g of zinc to produce H2 and ZnSO4.