Question

We wish to determine the moles of

carbon dioxide produced when
50.0 mL of 2.0 M hydrochloric acid
reacts with excess sodium
carbonate.

2HCl(aq) + Na₂CO3(aq) → 2NaCl(aq) + H₂O(1) + CO₂(g)

How many moles of HCI are
present
in 50.0 mL of 2.0 M HCI?

Answer 1

To determine the moles of HCl present in 50.0 mL of 2.0 M HCl, we can use the formula:

M = moles/volume

where M is the molarity of the solution, moles is the number of moles of solute, and volume is the volume of the solution in liters.

First, we need to convert the volume of the solution from milliliters to liters:

50.0 mL = 0.0500 L

Now we can use the formula to find the number of moles of HCl:

M = moles/volume
2.0 M = moles/0.0500 L
moles = 2.0 M * 0.0500 L
moles = 0.100 mol

Therefore, there are 0.100 moles of HCl present in 50.0 mL of 2.0 M HCl.

I hoped this helped!