What total volume (in lhter at 600 C and 1atm.) could be formed by the decomposition of 16g of NH4NO3 ?

see other post.

What is the molarity of a potassium hydroxide solution if 38.65 mL of the KOH solution is required to titrate 25.84 mL of 0.1982 M hydrochloric acid solution?

To find the total volume of gas formed by the decomposition of 16g of NH4NO3, we need to consider the balanced chemical equation for the decomposition reaction. The balanced equation is as follows:

NH4NO3(s) -> N2O(g) + 2H2O(g)

From the balanced equation, we can see that 1 mole of NH4NO3 decomposes to produce 1 mole of N2O gas and 2 moles of H2O gas.

First, let's calculate the number of moles of NH4NO3 from the given mass:

molar mass of NH4NO3 = 80 g/mol (14 + 4 + 14 + 3 * 16)
moles of NH4NO3 = mass / molar mass = 16g / 80 g/mol = 0.2 mol

Since 1 mole of NH4NO3 produces 1 mole of N2O gas, the moles of N2O gas formed will also be 0.2 mol.

Now, let's use the ideal gas law to find the volume of the N2O gas formed. The ideal gas law equation is:

PV = nRT

Where:
P = pressure = 1 atm
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

We need to convert the temperature of 600 C to Kelvin:

T(K) = 600 C + 273.15 = 873.15 K

Now, we can solve for the volume using the ideal gas law:

V = (nRT) / P
= (0.2 mol * 0.0821 L·atm/mol·K * 873.15 K) / 1 atm
≈ 14.32 L

Therefore, the total volume of gas formed by the decomposition of 16g of NH4NO3 at 600 C and 1 atm pressure is approximately 14.32 liters.