When 0.5963g of a sample containing NaNO2 and some inert material is allowed to react with excess HSO3NH2, the volume of N2 gas produced is measured. The volume of N2 gas corrected to 298.15K and 100kPa is calculated to be 0.146L. Find the mass percent of NaNO2 in the sample using the molar volume of N2 at 298.15K and 100kPa (from the previous question). The molar volume from the previous question was 22.8 L/mol.
To find the mass percent of NaNO2 in the sample, you need to determine the amount of NaNO2 that reacted and compare it to the total mass of the sample.
Step 1: Calculate the number of moles of N2 gas produced.
Using the ideal gas law equation, we can convert the volume of N2 gas produced to moles:
n = V / Vm
where n is the number of moles, V is the volume of the gas (0.146 L), and Vm is the molar volume of N2 at 298.15K and 100kPa (22.8 L/mol).
n = 0.146 L / 22.8 L/mol
Step 2: Calculate the number of moles of NaNO2.
Since NaNO2 is the reactant that produces N2 gas, the number of moles of N2 gas produced is equal to the number of moles of NaNO2 that reacted.
Step 3: Convert the moles of NaNO2 to grams.
To convert the moles of NaNO2 to grams, you need to multiply the number of moles by the molar mass of NaNO2.
The molar mass of NaNO2 can be calculated by adding up the atomic masses of sodium (Na), nitrogen (N), and oxygen (O).
Molar mass of NaNO2 = (22.98977 g/mol) + (14.00674 g/mol) + (2 * 15.999 g/mol)
Step 4: Calculate the mass percent of NaNO2.
The mass percent of NaNO2 in the sample can be calculated using the following formula:
Mass percent of NaNO2 = (mass of NaNO2 / total mass of the sample) x 100
Now you can substitute the values you calculated into the formula to find the mass percent of NaNO2.