Zinc reacts with hydrochloric acid according to the reaction equation Zn(s)+2HCL(aq)=>ZnCl2(aq)+H2(g). How many mm of 1.50M HCL(aq) are required to react with 9.05 g of ZN?
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = grams/atomic mass Zn
Convert mols Zn to mols HCl using the coefficients in the balanced equation.
Then convert mols HCl to volume using M = mols/L. You know mols and M, solv for L and convert to mL.
By the way, that is mL and not mm.
92.28
To find out how many milliliters (mm) of 1.50M HCl(aq) are needed to react with 9.05 g of Zn, we need to use stoichiometry to calculate the amount of HCl required. The stoichiometric ratio between Zn and HCl in the balanced equation is 1:2, which means that 1 mol of Zn reacts with 2 moles of HCl.
Step 1: Convert the mass of Zn to moles.
The molar mass of Zn is 65.38 g/mol.
Number of moles of Zn = mass / molar mass = 9.05 g / 65.38 g/mol = 0.1385 mol
Step 2: Determine the number of moles of HCl required.
Since the stoichiometric ratio is 1:2 (Zn:HCl), the number of moles of HCl required is twice the moles of Zn.
Number of moles of HCl = 2 * 0.1385 mol = 0.277 mol
Step 3: Calculate the volume of HCl solution.
The molarity of the HCl solution is given as 1.50M, which means 1.50 moles of HCl are present in 1 liter (1000 mL) of solution.
To find the volume of the solution containing 0.277 moles of HCl, we can use the formula:
Volume (in mL) = (Number of moles of HCl) / (Molarity of HCl) * 1000 mL/L
Volume (in mL) = (0.277 mol) / (1.50 mol/L) * 1000 mL/L = 184.67 mL
Therefore, approximately 184.67 milliliters of 1.50M HCl(aq) are required to react with 9.05 g of Zn.