For the following chemical reaction, what mass of silver nitrate in grams will be needed to produce 3.91 mol of calcium nitrate?
CaI2+2AgNO3=2AgI+Ca(NO3)2
Look at your PbO problem. This is done the same way but there is no percent part to the question. And you have mols initially so there is no conversion to start.
To determine the mass of silver nitrate (AgNO3) required to produce 3.91 mol of calcium nitrate (Ca(NO3)2), we need to use the balanced equation for the reaction:
CaI2 + 2AgNO3 → 2AgI + Ca(NO3)2
From the equation, we see that the stoichiometric ratio between calcium nitrate (Ca(NO3)2) and silver nitrate (AgNO3) is 1:2.
This means that for every 1 mole of calcium nitrate, we need 2 moles of silver nitrate.
First, we will calculate the number of moles of silver nitrate required:
Moles of AgNO3 = 2 × moles of Ca(NO3)2
= 2 × 3.91 mol
= 7.82 mol
Next, we need to convert the moles of silver nitrate into grams using the molar mass of AgNO3.
The molar mass of AgNO3 is:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3 atoms)
Molar mass of AgNO3:
= (107.87 g/mol) + (14.01 g/mol) + (16.00 g/mol x 3)
= 169.87 g/mol
To convert the moles of AgNO3 to grams:
Mass of AgNO3 = Moles of AgNO3 × Molar mass of AgNO3
= 7.82 mol × 169.87 g/mol
= 1324.12 g
Therefore, approximately 1324.12 grams of silver nitrate will be needed to produce 3.91 mol of calcium nitrate.
To find the mass of silver nitrate (AgNO3) needed to produce 3.91 moles of calcium nitrate (Ca(NO3)2), we need to use the stoichiometry of the balanced chemical equation.
The balanced equation shows that 2 moles of silver nitrate react with 1 mole of calcium iodide to produce 2 moles of silver iodide and 1 mole of calcium nitrate.
Based on the stoichiometry, we can set up the following ratio:
2 moles AgNO3 : 1 mole Ca(NO3)2
To convert the given moles of Ca(NO3)2 to the corresponding moles of AgNO3, we multiply the given moles by the ratio:
3.91 moles Ca(NO3)2 × (2 moles AgNO3 / 1 mole Ca(NO3)2) = 7.82 moles AgNO3
Now, we need to convert the moles of silver nitrate to grams using its molar mass.
The molar mass of silver nitrate (AgNO3) can be calculated by adding the atomic masses of each element:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (3 oxygen atoms)
Molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + (16.00 g/mol × 3) = 169.87 g/mol
Finally, we can calculate the mass of silver nitrate required:
Mass of AgNO3 = moles of AgNO3 × molar mass of AgNO3
Mass of AgNO3 = 7.82 moles × 169.87 g/mol = 1327.89 grams (rounded to four decimal places)
Therefore, approximately 1327.89 grams of silver nitrate will be needed to produce 3.91 moles of calcium nitrate.