I am working on a redox stoichiometry lab. I need to calculate the moles NaBO3*4H2O.
Mass=1.0133g in 100mL solution
In my experiment I used 9.98mL and titrated in with KMnO4.
I'm really confused about how to calculate the moles. Thanks for any help!!!!
Write the equation and balance it. Do that first.
I calculated the M of NaBO.... at 6.59x10^-3 for the mass I measured. I have a chart of measurements where I used 9.98mL out of 100mL of solution that I made. I was given the equation
(mass Na.../molar mass Na...)*(volume Na.../volume total solution)
the balanced equation I'm totally stuck.
I need more information than that. For moles NaBO3.3H2O = grams/molar mass.
To calculate the moles of NaBO3*4H2O, you can follow these steps:
1. Determine the molar mass of NaBO3*4H2O:
- Molar mass of Na = 22.99 g/mol
- Molar mass of B = 10.81 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of H = 1.01 g/mol
- Adding all the molar masses together: (22.99 + 10.81 + (3 * 16.00) + (8 * 1.01)) g/mol = 201.96 g/mol.
2. Calculate the moles of NaBO3*4H2O using the given mass:
- Mass NaBO3*4H2O = 1.0133 g (given)
- Moles NaBO3*4H2O = Mass NaBO3*4H2O / Molar mass NaBO3*4H2O
= 1.0133 g / 201.96 g/mol
≈ 0.00502 mol.
So, the moles of NaBO3*4H2O is approximately 0.00502 mol for the given mass.
Note: Make sure to check if your solution contains any impurities or additional compounds that may affect the stoichiometry.