Find the number of mL of 4M HNO3 solution required to react with 216g of Ag according to the equation 3Ag+4HNO3->3AgNO3+2H2O

To find the volume of 4M HNO3 solution required, we can use the stoichiometry of the balanced equation.

Step 1: Calculate the moles of Ag.
Given mass of Ag = 216g
Molar mass of Ag = 108g/mol
Number of moles of Ag = mass / molar mass = 216g / 108g/mol = 2 moles

Step 2: Determine the ratio between Ag and HNO3.
From the balanced equation, we see that 3 moles of Ag react with 4 moles of HNO3.

Step 3: Calculate the moles of HNO3 needed.
Using the ratio of 3 moles of Ag to 4 moles of HNO3, we can set up a proportion:
3 moles Ag / 4 moles HNO3 = 2 moles Ag / x
Cross-multiplying, we get:
3 moles Ag * x = 2 moles Ag * 4 moles HNO3
3x = 8
x = 8 / 3

Step 4: Convert moles of HNO3 to volume.
We know that the concentration of HNO3 is 4M, which means there are 4 moles of HNO3 in 1 liter of solution (1000 mL).
Using this information, we multiply the moles of HNO3 by the volume of 1 mole of HNO3 (1000 mL / 4 moles):
x mL = (8 / 3) * (1000 mL / 4 moles)
x mL = 2000 / 3 mL
x mL ≈ 666.67 mL (rounded to two decimal places)

Therefore, approximately 666.67 mL of 4M HNO3 solution is required to react with 216g of Ag.

To find the number of mL of 4M HNO3 solution required to react with 216g of Ag, we first need to calculate the number of moles of Ag using its molar mass and then use the stoichiometry of the balanced chemical equation to determine the number of moles of HNO3 required.

1. Calculate the number of moles of Ag:
- Molar mass of Ag = 108 g/mol
- Moles of Ag = Mass of Ag / Molar mass of Ag
= 216 g / 108 g/mol
= 2 moles

2. Now we use the stoichiometry of the balanced equation to determine the number of moles of HNO3 required.
According to the balanced equation: 3Ag + 4HNO3 -> 3AgNO3 + 2H2O
The stoichiometric ratio of Ag to HNO3 is 3:4.

- Moles of HNO3 = (2 moles Ag) x (4 moles HNO3 / 3 moles Ag)
= 8/3 moles

3. Calculate the volume of 4M HNO3 required based on the given 4M concentration.
- Molarity (M) = Moles / Volume (in liters)
- Volume (in liters) = Moles / Molarity
- Volume (in mL) = Volume (in liters) x 1000

- Volume (in liters) = (8/3 moles HNO3) / 4 M
= (8/3) / 4 L
= 2/3 L
= 0.67 L

- Volume (in mL) = 0.67 L x 1000
= 670 mL

Therefore, 670 mL of the 4M HNO3 solution is required to react with 216g of Ag.