Use the reaction below to determine the volume of 0.123 M AgNO3(aq)that is needed to form 0.657 g of Ag2SO4 (s).
2AgNO3(aq) + H2SO4(aq) �¨ Ag2SO4 (s) + 2HNO3(l)
2AgNO3 + H2SO4-----Ag2SO4 + 2HNO3
Here is a worked example of a stoichiometry problem. http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of 0.123 M AgNO3(aq) needed to form 0.657 g of Ag2SO4(s), we first need to calculate the number of moles of Ag2SO4 formed.
Step 1: Calculate the molar mass of Ag2SO4.
Ag2SO4 has a molar mass of:
(2 × atomic mass of Ag) + atomic mass of S + (4 × atomic mass of O)
We can find the atomic masses of Ag, S, and O from the periodic table. Adding them up, we get:
(2 × 107.87 g/mol) + 32.06 g/mol + (4 × 16.00 g/mol) = 231.14 g/mol
Step 2: Calculate the number of moles of Ag2SO4 using the formula:
moles = mass / molar mass
Plugging in the values, we have:
moles = 0.657 g / 231.14 g/mol
Calculating this, we get the number of moles of Ag2SO4.
Step 3: Apply the stoichiometry of the reaction to determine the number of moles of AgNO3.
Looking at the balanced chemical equation:
2 AgNO3 + H2SO4 → Ag2SO4 + 2 HNO3
From the equation, we can see that 2 moles of AgNO3 are needed to produce 1 mole of Ag2SO4. So, we multiply the number of moles of Ag2SO4 by the stoichiometric ratio:
moles of AgNO3 = (1 mole of Ag2SO4) × (2 moles of AgNO3 / 1 mole of Ag2SO4)
Step 4: Calculate the volume of 0.123 M AgNO3(aq) using the formula:
volume = moles / concentration
The concentration is given as 0.123 M, which means that for every 1 liter of the AgNO3 solution, there are 0.123 moles of AgNO3. So,
volume = moles of AgNO3 / concentration
Plug in the calculated value for moles of AgNO3 and the provided concentration to find the volume of AgNO3 needed to form Ag2SO4.