Solid ammonium nitrate breaks down into nitrogen, water vapor, and oxygen. If there was 300g of the solid and the reaction happened under STP find the total volume of the products. (translating words to eqns., balancing eqns., understanding what STP means) thanks!
P.S. eqns.=equations
2NH4NO3 ==> 2N2 + 4H2O + O2
STP means standard temperature and pressure.
Convert 300 g NH4NO3 to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles NH4NO3 to moles N2, then to moles H2O, then to moles O2.
Finally, convert moles N2, H2O, and O2 to liters. L = moles x 22.4 L/mol at STP. Then add the L to obtain total gas volume.
Check the equation to make sure it is balanced. Post your work if you get stuck.
To find the total volume of the products, we first need to determine the balanced equation for the reaction. From the given information, we know that solid ammonium nitrate (NH4NO3) decomposes into nitrogen (N2), water vapor (H2O), and oxygen (O2).
The balanced equation for this reaction is:
2NH4NO3(s) → 2N2(g) + O2(g) + 4H2O(g)
Next, let's calculate the molar mass of ammonium nitrate (NH4NO3) to determine the number of moles present in 300 grams. The molar mass of NH4NO3 is:
N - 14.01 g/mol
H - 1.01 g/mol (4 hydrogen atoms)
O - 16.00 g/mol (3 oxygen atoms)
Adding these together:
14.01 + (1.01 * 4) + (16.00 * 3) = 80.05 g/mol
Therefore, the number of moles of solid ammonium nitrate is:
300 g / 80.05 g/mol = 3.75 moles of NH4NO3
Since the reaction occurs at STP (Standard Temperature and Pressure), we can assume that 1 mole of any gas occupies 22.4 liters at a temperature of 273 K and a pressure of 1 atmosphere.
According to the balanced equation, 2 moles of NH4NO3 produce 2 moles of N2, 1 mole of O2, and 4 moles of H2O. Thus, 3.75 moles of NH4NO3 will produce:
3.75 moles * (2 moles/ 2 moles) = 3.75 moles of N2
3.75 moles * (1 mole/ 2 moles) = 1.875 moles of O2
3.75 moles * (4 moles/ 2 moles) = 7.5 moles of H2O
Using the ideal gas law, we can now calculate the total volume of the products:
Volume of N2 = 3.75 moles * 22.4 L/mol = 84 L
Volume of O2 = 1.875 moles * 22.4 L/mol = 42 L
Volume of H2O = 7.5 moles * 22.4 L/mol = 168 L
Therefore, the total volume of the products (N2, O2, and H2O) is:
84 L + 42 L + 168 L = 294 L