How many milliliters of 3.0 M NaOH are required to react with 4.0mL of 16.0 M HNO3? ---> Not sure if I am solving correctly but I keep getting 21mL.
21.333 mL which rounds to 21 to 2 s.f.
To solve this problem, you need to use the concept of stoichiometry and the balanced chemical equation for the reaction between NaOH and HNO3. The balanced equation is:
NaOH + HNO3 → NaNO3 + H2O
Here's how you can calculate the number of milliliters of 3.0 M NaOH required to react with 4.0 mL of 16.0 M HNO3:
1. Determine the number of moles of HNO3:
Volume(HNO3) x Concentration(HNO3) = Moles(HNO3)
4.0 mL x 16.0 mol/L = 64.0 mmol (millimoles)
2. Use the stoichiometry of the balanced equation to determine the amount of NaOH required. The stoichiometry tells us that 1 mole of NaOH reacts with 1 mole of HNO3.
Moles(NaOH) = Moles(HNO3) = 64.0 mmol
3. To convert moles to milliliters, you can use the molarity (concentration) of NaOH.
Moles(NaOH) = Volume(NaOH) x Concentration(NaOH)
64.0 mmol = Volume(NaOH) x 3.0 mol/L
4. Rearrange the equation to solve for the volume of NaOH:
Volume(NaOH) = (64.0 mmol) / (3.0 mol/L)
Volume(NaOH) = 21.3 mL (approximately)
Therefore, approximately 21.3 mL of 3.0 M NaOH is required to react with 4.0 mL of 16.0 M HNO3.
To determine the quantity of NaOH needed to react with a given volume of HNO3, we can use the concept of stoichiometry. The balanced equation for the reaction between NaOH and HNO3 is:
2NaOH + HNO3 -> NaNO3 + 2H2O
From the equation, we can see that it takes 2 moles of NaOH to react with 1 mole of HNO3.
First, let's calculate the number of moles of HNO3 in 4.0 mL of 16.0 M HNO3:
Moles of HNO3 = (concentration of HNO3) x (volume of HNO3)
= 16.0 M x (4.0 mL ÷ 1000 mL/mL) [converting mL to L]
= 0.064 mol
Since the stoichiometry of the reaction is 2:1 (NaOH to HNO3), we need twice as many moles of NaOH to react with the HNO3. Therefore, the number of moles of NaOH required is:
Moles of NaOH = 2 x (moles of HNO3)
= 2 x 0.064 mol
= 0.128 mol
Now, let's calculate the volume of 3.0 M NaOH needed to obtain 0.128 mol of NaOH:
Volume of NaOH = (moles of NaOH) / (concentration of NaOH)
= 0.128 mol / 3.0 M
= 0.0427 L or 42.7 mL [rounded to 3 significant figures]
Therefore, the correct answer is approximately 42.7 mL of 3.0 M NaOH, not 21 mL.
Please note that you should always double-check your calculations and make sure you're using the correct stoichiometry when solving these types of problems.