An aqueous solution of ammonium nitrate NH4NO2 decomposeds when heated to give off nitrogen.

This reaction may be used to prepare pure nitrogen. How many grams of ammonium nitrite must have reacted if 3.75dm^3 of nitrogen gas was collected over water at 26 degrees C and 97.8 kPa?

Same with this one can some one help me start it out. please

Write the equation and balance it.

Convert 3.75 dm^3 to L. Convert kPa to atm. Use PV = nRT and solve for n. Convert mols N2 to mols NH4NO3 and that to grams.

To solve this problem, you will need to follow a few steps:

Step 1: Write the balanced chemical equation for the decomposition of ammonium nitrite (NH4NO2) into nitrogen (N2) gas:
NH4NO2 → N2 + 2H2O

Step 2: Convert the given volume of nitrogen gas (in dm^3) to moles. To do this, you can use the ideal gas law equation:
PV = nRT

Where:
P = pressure in atm (97.8 kPa can be converted to atm by dividing by 101.325)
V = volume in liters (3.75 dm^3)
n = number of moles (what we want to find)
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature in Kelvin (26°C can be converted to Kelvin by adding 273)

Step 3: Use the stoichiometric coefficients from the balanced equation to find the number of moles of ammonium nitrite that reacted. In the balanced equation, the coefficient in front of NH4NO2 is 1.

Step 4: Convert the number of moles of ammonium nitrite to grams using the molar mass of NH4NO2. The molar mass can be calculated by adding up the atomic masses of each element in the compound (N = 14.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol).

By following these steps, you will be able to determine the number of grams of ammonium nitrite that reacted.