100.0cm3 of 2.0mol dm-3 hydrochloric acid is mixed with 2.5g of magnesium oxide powder. The mixture is stirred until no further change occurs. The resulting solution still contains excess hydrochloric acid. Calculate the volume of 1.5mol dm -3 sodium hydroxide solution needed to neutralise the resulting solution.(Relative atomic mass:O=16; Mg=24)
MgO + 2HCl => H2O + MgCl2
mols MgO = grams/molar mass = ?
mols HCl = M x L = ?
The problem tells you that all of the MgO was used but not all of the HCl was used.
How much HCl was used. That's mols MgO x (2 mols HCl/1 mol MgO) = ? mols HCl.
How much HCl is left. Subtract initial mols HCl from mols HCl used. that's what's left. That will be neutralized with the NaOH as follows.
NaOH + HCl ==> H2O + NaCl
Moles NaOH needed = mols HCl you have in the final solution.
Then M = mols NaOH/L NaOH
You know M NaOH (M = mols/dm^3), you know mols, solve for L NaOH. Post your work if you get stuck.
To answer this question, we need to calculate the number of moles of hydrochloric acid initially present in the solution, then determine the number of moles of hydrochloric acid remaining after it reacts with the magnesium oxide. Finally, we will calculate the volume of sodium hydroxide solution needed to neutralize the remaining hydrochloric acid.
Let's break down the steps:
Step 1: Calculate the number of moles of hydrochloric acid initially present:
- We are given the concentration of hydrochloric acid as 2.0 mol dm^-3 and the volume as 100.0 cm^3.
- To convert from cm^3 to dm^3, we divide the volume by 1000.
- The equation for calculating moles is: moles = concentration (mol dm^-3) * volume (dm^3).
- Plugging in the values, we get: moles of HCl = 2.0 mol dm^-3 * (100.0 cm^3 / 1000) dm^3.
Step 2: Calculate the number of moles of magnesium oxide:
- We are given the mass of magnesium oxide as 2.5 g.
- To calculate moles, we divide the mass by the molar mass of magnesium oxide.
- The molar mass of magnesium oxide (MgO) is the sum of the atomic masses of Mg and O.
- Plugging in the atomic masses (Mg = 24 g/mol, O = 16 g/mol), we get: moles of MgO = 2.5 g / (24 g/mol + 16 g/mol).
Step 3: Determine the limiting reactant:
- To find the limiting reactant, we compare the number of moles of hydrochloric acid with the number of moles of magnesium oxide.
- The reactant that has fewer moles is the limiting reactant. It determines the amount of product formed.
- In this case, whichever has fewer moles, HCl or MgO, will run out first and limit the reaction.
- We compare the moles of HCl calculated in Step 1 with the moles of MgO calculated in Step 2.
Step 4: Calculate the moles of hydrochloric acid remaining:
- Since hydrochloric acid is in excess, we need to find how much is left after it reacts with magnesium oxide.
- To do that, we subtract the moles of magnesium oxide used from the moles of hydrochloric acid initially present.
- This gives us the moles of hydrochloric acid remaining.
Step 5: Calculate the volume of sodium hydroxide solution needed to neutralize the remaining hydrochloric acid:
- Since we know the concentration of the sodium hydroxide solution (1.5 mol dm^-3), we can use the equation: moles = concentration (mol dm^-3) * volume (dm^3).
- We plug in the moles of hydrochloric acid remaining and the concentration of sodium hydroxide to find the volume needed.
Following these steps, we can calculate the volume of sodium hydroxide solution needed to neutralize the remaining hydrochloric acid in the resulting mixture.