what mass of sodium chloride will be required to completely react with 4.00 x 10(24) units of AgNO3 and form AgCl and sodium nitrate?
To determine the mass of sodium chloride required to completely react with 4.00 x 10^24 units of AgNO3, you will need to use stoichiometry.
1. Write the balanced chemical equation for the reaction:
2 AgNO3 + NaCl -> 2 AgCl + NaNO3
2. Determine the molar ratio between AgNO3 and NaCl:
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of NaCl.
3. Convert the number of units of AgNO3 to moles:
Given that there are 4.00 x 10^24 units of AgNO3, we need to convert this to moles. The molar mass of AgNO3 is 169.87 g/mol.
4. Calculate the moles of AgNO3:
moles of AgNO3 = (4.00 x 10^24 units) / (6.022 x 10^23 units/mol) = 6.64 moles
5. Convert moles of AgNO3 to moles of NaCl:
Since the molar ratio between AgNO3 and NaCl is 2:1, the number of moles of NaCl will be half of the moles of AgNO3.
moles of NaCl = (6.64 moles of AgNO3) / 2 = 3.32 moles
6. Calculate the mass of NaCl:
To calculate the mass, we need to know the molar mass of NaCl, which is 58.44 g/mol.
mass of NaCl = (3.32 moles) x (58.44 g/mol) = 193.46 grams
Therefore, the mass of sodium chloride required to completely react with 4.00 x 10^24 units of AgNO3 and form AgCl and sodium nitrate is 193.46 grams.
To determine the mass of sodium chloride required to react with a given amount of silver nitrate, we need to use the balanced chemical equation for the reaction and convert the given units to moles. From the balanced equation:
AgNO3 + NaCl → AgCl + NaNO3
We can see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to form 1 mole of silver chloride and 1 mole of sodium nitrate.
Given:
Amount of AgNO3 = 4.00 x 10^24 units
Step 1: Convert the given units of AgNO3 to moles.
To do this, we need to know the molar mass of AgNO3. AgNO3 consists of one silver atom (Ag), one nitrogen atom (N), and three oxygen atoms (O).
The molar masses are:
AgNO3 = 107.87 g/mol (Ag) + 14.01 g/mol (N) + 3(16.00 g/mol) (O) = 169.87 g/mol
Now, we can calculate the number of moles of AgNO3:
moles of AgNO3 = given amount / molar mass
moles of AgNO3 = (4.00 x 10^24 units) / (6.022 x 10^23 units/mol)
moles of AgNO3 = 6.64 moles
Step 2: Determine the moles of NaCl required to react with AgNO3.
Since the reaction is 1:1 stoichiometry (1 mole of AgNO3 reacts with 1 mole of NaCl), the moles of NaCl required will also be 6.64 moles.
Step 3: Convert the moles of NaCl to grams.
To do this, we need to know the molar mass of NaCl. NaCl consists of one sodium atom (Na) and one chlorine atom (Cl).
The molar masses are:
NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol
Now, we can calculate the mass of NaCl:
mass of NaCl = moles of NaCl × molar mass
mass of NaCl = 6.64 moles × 58.44 g/mol
mass of NaCl = 388.10 g
Therefore, approximately 388.10 grams of sodium chloride will be required to completely react with 4.00 x 10^24 units of silver nitrate and form silver chloride and sodium nitrate.
Here is a worked example of a stoichiometry problem. Just follow the steps. I don't think the example shows how to get to mols from molecules.
mols AgNO3 = 4E24/6.02E23 = ?
The example will take this value and go to the end. This and the limiting reagent process I gave you last yesterday will work a bunch of problems. I suggest you print them out for future reference.
Balanced equation is:
NaCl + AgnO3 yields AgCl + NaNO3
1mol 1 mol 1 mol 1 mol
NaCl= 58.44
This is as far as I get. My answer choices are:
6.64g
0.144g
388g
8.80g