how many mL of 0.50 M ammonia would be needed to react with 1.0 L of 0.75 M HClO4

How many moles HClO4 do you have? M x L = moles.

How many mols NH3 will that require. Write the equation and you will see 1 mol HClO4 requires just 1 mol NH3; therefore, moles HClO4 = mole NH3.
Now M NH3 = moles NH3/L NH3. Solve for L and convert to mL.

To determine the volume of 0.50 M ammonia needed to react with 1.0 L of 0.75 M HClO4, we need to use the concept of stoichiometry and the balanced chemical equation.

First, let's write the balanced equation for the reaction between ammonia (NH3) and perchloric acid (HClO4):

NH3 + HClO4 → NH4ClO4

From the balanced equation, we can see that the stoichiometric ratio between ammonia and perchloric acid is 1:1. This means that for every 1 mole of NH3, we need 1 mole of HClO4 to react completely.

To find out the number of moles of HClO4 in 1.0 L of 0.75 M HClO4, we can use the formula:

moles = concentration (M) × volume (L)

moles of HClO4 = 0.75 M × 1.0 L = 0.75 moles

Since the stoichiometric ratio between NH3 and HClO4 is 1:1, we would also need 0.75 moles of NH3 to react completely.

Now, let's calculate the volume of 0.50 M ammonia required using the formula:

volume (L) = moles / concentration (M)

volume of NH3 = 0.75 moles / 0.50 M = 1.5 L

Therefore, 1.5 L of 0.50 M ammonia would be needed to react with 1.0 L of 0.75 M HClO4.