How many milliliters of 6.50 M HCl(aq) are required to react with 5.95 g of an ore containing 35.0% Zn(s) by mass?
Zn + 2HCl ==> Zn^2+ ....
g Zn = 5.95*0.35 = ?
mols Zn = ?g Zn/atomic mass Zn
Convert mols Zn to mols HCl using the coefficients.
M HCl = mols HCl/L HCl. You know M and mols, solve for L and convert to mL.
To determine the number of milliliters of 6.50 M HCl(aq) required to react with 5.95 g of an ore containing 35.0% Zn(s) by mass, you will need to follow these steps:
1. Calculate the mass of Zn(s) in the ore:
Mass of Zn(s) = 5.95 g * 0.350 = 2.0825 g
2. Convert the mass of Zn(s) to moles:
Moles of Zn(s) = Mass of Zn(s) / molar mass of Zn(s)
The molar mass of Zn(s) can be found on the periodic table and is equal to 65.38 g/mol.
Moles of Zn(s) = 2.0825 g / 65.38 g/mol = 0.0318 mol
3. Since the balanced chemical equation for the reaction is not given, we will assume it is:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
From the balanced chemical equation, we know that 1 mole of Zn(s) reacts with 2 moles of HCl(aq). Therefore, the moles of HCl(aq) required can be calculated using the mole ratio:
Moles of HCl(aq) = 0.0318 mol Zn(s) * (2 mol HCl(aq) / 1 mol Zn(s)) = 0.0636 mol
4. Finally, we can calculate the volume of 6.50 M HCl(aq) required using the molarity equation:
Moles of solute = Molarity * Volume (in liters)
Rearranging the equation to solve for the volume, we have:
Volume (in liters) = Moles of solute / Molarity
Volume (in liters) = 0.0636 mol / 6.50 M = 0.0098 L
5. Convert the volume from liters to milliliters by multiplying by 1000:
Volume (in milliliters) = 0.0098 L * 1000 = 9.8 mL
Therefore, approximately 9.8 milliliters of 6.50 M HCl(aq) are required to react with 5.95 g of an ore containing 35.0% Zn(s) by mass.