Silver metal react with nitric acid according to the equation
3Ag+4HNO3----->3AgNO3+NO+H2O
The volume 1.5 M of HNO3 required react with 0.784gm silver is
how many moles of Ag in .784g ?
You will need 4/3 that many moles of HNO3
How many liters contain that many moles?
To determine the volume of 1.5 M HNO3 required to react with 0.784 g of silver, we need to use stoichiometry.
Step 1: Determine the moles of silver (Ag):
To do this, divide the given mass of silver by its molar mass. The molar mass of silver (Ag) is 107.87 g/mol.
Moles of Ag = Mass of Ag / Molar Mass of Ag
= 0.784 g / 107.87 g/mol
= 0.00725 mol
Step 2: Use the stoichiometric ratio to find the moles of HNO3:
From the balanced equation, the stoichiometric ratio of Ag to HNO3 is 3:4.
Moles of HNO3 = (Moles of Ag) × (4 moles HNO3 / 3 moles Ag)
= 0.00725 mol × (4/3)
= 0.00967 mol
Step 3: Determine the volume of 1.5 M HNO3 required:
The concentration of HNO3 is given as 1.5 M, which means there are 1.5 moles of HNO3 dissolved in 1 liter of solution.
Volume of HNO3 = Moles of HNO3 / Concentration of HNO3
= 0.00967 mol / 1.5 mol/L
= 0.00645 L
Since 1 L is equal to 1000 mL, we can convert liters to milliliters:
Volume of HNO3 = 0.00645 L × 1000 mL/L
= 6.45 mL
Therefore, the volume of 1.5 M HNO3 required to react with 0.784 g of silver is approximately 6.45 mL.
To find the volume of 1.5 M nitric acid (HNO3) that is required to react with 0.784 grams of silver (Ag), we first need to determine the amount of silver in moles.
Step 1: Calculate the number of moles of silver
To find the number of moles, you can use the formula:
moles = mass / molar mass
The molar mass of silver (Ag) can be found on the periodic table, which is approximately 107.87 g/mol.
moles of Ag = 0.784 g / 107.87 g/mol
moles of Ag ≈ 0.00726 mol
Step 2: Determine the stoichiometry of the reaction
According to the balanced chemical equation:
3Ag + 4HNO3 → 3AgNO3 + NO + H2O
The stoichiometry of the reaction tells us that 3 moles of silver (Ag) react with 4 moles of nitric acid (HNO3). Therefore, the ratio of moles of Ag to moles of HNO3 is 3:4.
Step 3: Calculate the moles of HNO3 required
Since the ratio of HNO3 to Ag is 4:3, we can set up a proportion using the moles of Ag we calculated earlier:
moles of HNO3 = (moles of Ag) x (moles of HNO3 / moles of Ag)
moles of HNO3 = 0.00726 mol x (4 mol HNO3 / 3 mol Ag)
moles of HNO3 ≈ 0.00968 mol
Step 4: Calculate the volume of 1.5 M HNO3 required
The concentration of the nitric acid given is 1.5 M, which means there are 1.5 moles of HNO3 in 1 liter (1000 mL) of solution.
volume of HNO3 = (moles of HNO3) / (concentration of HNO3)
volume of HNO3 = 0.00968 mol / 1.5 M
volume of HNO3 ≈ 0.00645 L or 6.45 mL
Therefore, approximately 6.45 mL of 1.5 M nitric acid is required to react with 0.784 grams of silver.