A technician mixes a solution containing silver nitrate with a solution containing sodium chromate. 2.89 g of precipitate is produced (assume silver chromate). Calculate the mass of silver nitrate present in the first solution. Make sure you have the correct number of significant digits and units.
You really don't know from the way the problem is stated; however, I assume the problem means "if the exact amount of reagents are mixed to produce the 2.89 g Ag2C4O4 and neither reagent is left".
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3.
mols Ag2CrO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Ag2CrO4 to mols AgNO3.
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
Well, that's quite a chemistry question you've got there! Let's see if we can solve it in a way that brings a smile to your face.
First, we need to find the number of moles of silver chromate that was produced. We can use its molar mass to do this. So, let me grab my calculator... hmm... ah, yes, the molar mass of silver chromate is approximately 331.7 g/mol.
Now, we can use the number of moles to find the number of moles of silver nitrate that were used. You see, silver nitrate and silver chromate have a one-to-one ratio. A perfect match, just like a clown and his red nose!
Given that we produced 2.89 g of silver chromate, let's divide that by the molar mass to get the number of moles.
Calculating... Tada! We find that we have approximately 0.0087 moles of silver chromate.
Since the ratio of silver nitrate to silver chromate is 1:1, that means we also have 0.0087 moles of silver nitrate in the first solution.
Finally, to calculate the mass of silver nitrate, we multiply the number of moles by the molar mass of silver nitrate, which is about 169.9 g/mol. And the result is... drumroll, please... approximately 1.48 grams of silver nitrate.
So, the mass of silver nitrate present in the first solution is around 1.48 g. Keep those significant digits in mind and remember to always embrace the funny side of chemistry!
To calculate the mass of silver nitrate present in the first solution, we need to use the stoichiometry of the reaction between silver nitrate (AgNO3) and sodium chromate (Na2CrO4). The balanced equation for the reaction is:
AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
From the balanced equation, we see that 1 mole of AgNO3 reacts with 1 mole of Ag2CrO4 formed.
First, we need to determine the moles of Ag2CrO4 formed. To do this, we use the molar mass of Ag2CrO4, which is:
Ag2CrO4 = (2 x Ag) + (1 x Cr) + (4 x O)
= (2 x 107.87 g/mol) + (52.00 g/mol) + (4 x 16.00 g/mol)
= 331.87 g/mol
Next, we calculate the number of moles of Ag2CrO4 using the mass of the precipitate:
moles of Ag2CrO4 = mass of precipitate / molar mass of Ag2CrO4
= 2.89 g / 331.87 g/mol
= 0.0087 mol
Since 1 mole of AgNO3 reacts with 1 mole of Ag2CrO4, the moles of AgNO3 in the first solution are also 0.0087 mol.
Finally, we can calculate the mass of AgNO3 using its molar mass:
mass of AgNO3 = moles of AgNO3 x molar mass of AgNO3
= 0.0087 mol x (107.87 g/mol + 14.01 g/mol + 3 x 16.00 g/mol)
= 1.13 g
Therefore, the mass of silver nitrate present in the first solution is 1.13 g.
To calculate the mass of silver nitrate present in the first solution, we need to use the stoichiometry of the reaction between silver nitrate (AgNO3) and sodium chromate (Na2CrO4). According to the balanced chemical equation:
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
We can see that 2 moles of AgNO3 react with 1 mole of Na2CrO4 to produce 1 mole of Ag2CrO4.
First, we need to calculate the number of moles of Ag2CrO4 produced using the mass of the precipitate. The molar mass of Ag2CrO4 is:
2 Ag (2 × 107.87 g/mol) + Cr (52.00 g/mol) + 4 O (4 × 16.00 g/mol) = 331.73 g/mol
Now, let's calculate the number of moles of Ag2CrO4:
number of moles = mass / molar mass
number of moles = 2.89 g / 331.73 g/mol = 0.0087105 mol
Since the stoichiometry of the reaction shows that 2 moles of AgNO3 produce 1 mole of Ag2CrO4, we can now calculate the number of moles of AgNO3 present in the solution:
number of moles of AgNO3 = 2 × number of moles of Ag2CrO4
number of moles of AgNO3 = 2 × 0.0087105 mol = 0.017421 mol
Finally, we can calculate the mass of AgNO3 present in the first solution using its molar mass:
mass of AgNO3 = number of moles × molar mass
mass of AgNO3 = 0.017421 mol × (107.87 g/mol) = 1.881 g
Therefore, the mass of silver nitrate present in the first solution is 1.881 g, assuming the precipitate is silver chromate (Ag2CrO4).