A 50g sample of ammonium dichromate is decomposed, what is the milliliter volume of nitrogen gas released at STP?

the balance equation;

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)

50g of ammonium dichromate is 0.198 mole (50/252.07). according to the equation, 1mole of ammonium dichromate produces 1 mole of nitrogen gas, so 0.198mole of dichromate produces 0.198mole of nitrogen gas.

mass of nitrogen gas = 0.198molex28g/mol = 5.54g

use the density of nitrogen gas to find the volume...

hope that helps..

if not, then recall that for any ideal gas at STP, 1mol=22.4L. and already we have the mole for nitrogen (0.198mole). so we can find the volume;

1mol=22.4L
0.198mol=x

x = 4.435L = 4435mL of Nitrogen gas.

To determine the volume of nitrogen gas released at STP during the decomposition of ammonium dichromate, we need to calculate the number of moles of nitrogen gas produced and then use the ideal gas law to find its volume.

1. Calculate the number of moles of ammonium dichromate:
Since we have a 50g sample, we need to determine the molar mass of ammonium dichromate (NH4)2Cr2O7. By adding the atomic masses of each element (2x atomic mass of N + 8x atomic mass of H + 2x atomic mass of Cr + 7x atomic mass of O), we get:
(2x14.01 g/mol) + (8x1.01 g/mol) + (2x52.00 g/mol) + (7x16.00 g/mol) = 252.04 g/mol

Now we can calculate the number of moles of ammonium dichromate:
50g / 252.04 g/mol = 0.1984 mol

2. Determine the balanced chemical equation for the decomposition of ammonium dichromate:
(NH4)2Cr2O7 → Cr2O3 + N2 + 4H2O

From the balanced equation, we see that 1 mole of ammonium dichromate produces 1 mole of nitrogen gas.

3. Use the ideal gas law to calculate the volume of nitrogen gas at STP (Standard Temperature and Pressure):
At STP, the conditions are 273.15 K (0°C) and 1 atmosphere of pressure.

The ideal gas law equation is:
PV = nRT

Since we want to find the volume (V), we rearrange the equation to:
V = nRT / P

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

Plugging in the values:
V = (0.1984 mol) x (0.0821 L·atm/(mol·K)) x (273.15 K) / (1 atm)

Calculating this yields:
V ≈ 4.28 L

Therefore, approximately 4.28 liters of nitrogen gas would be released at STP during the decomposition of a 50g sample of ammonium dichromate.