4{C3H5(NO3)2}(l)--->6N2(g)+O2(g)+12CO2(g)+10H2O(g)
If a dynamite of 100mL needs 3.5atm to explode, what minimum mass of
C3H5(NO3)2 do I need to put in at 25 degree celcius?
I tried to do this by finding the number of moles by the formula PV=nRT and then convert that into masss.is this correct?
if not please explain why.
thanks in advance!
To determine the minimum mass of C3H5(NO3)2 required for the dynamite to explode, you will indeed need to find the number of moles first. However, the formula PV = nRT alone won't give you the number of moles of C3H5(NO3)2.
The equation you provided indicates that the reaction involves 4 moles of C3H5(NO3)2. So, you need to calculate the number of moles of N2 gas produced and use that to determine the number of moles of C3H5(NO3)2.
Here's a step-by-step explanation of how to solve the problem:
Step 1: Convert the volume of the dynamite from mL to liters:
100 mL = 0.1 L
Step 2: Convert the pressure from atm to Pa:
1 atm = 101,325 Pa
Step 3: Use the ideal gas law equation, PV = nRT, to find the number of moles of N2 gas:
n = PV / RT
Given:
P = 3.5 atm (converted to Pa)
V = 0.1 L
R = 8.314 J/(mol·K) (universal gas constant)
T = 25 °C = 25 + 273 = 298 K
Substituting the values into the equation:
n = (3.5 atm * 101,325 Pa/atm * 0.1 L) / (8.314 J/(mol·K) * 298 K)
n ≈ 1.4876 moles of N2 gas
Step 4: Determine the number of moles of C3H5(NO3)2:
From the balanced equation, we can see that for every 6 moles of N2 gas produced, 4 moles of C3H5(NO3)2 are required.
Using a proportion, we can set up the following equation:
6 moles N2 = 4 moles C3H5(NO3)2
1.4876 moles N2 = x moles C3H5(NO3)2
Solving for x:
x = (1.4876 mol C3H5(NO3)2 * 4 mol C3H5(NO3)2) / 6 mol N2
x ≈ 0.9917 moles of C3H5(NO3)2
Step 5: Calculate the molar mass of C3H5(NO3)2:
C = 3 * 12.01 g/mol
H = 5 * 1.008 g/mol
N = 2 * 14.01 g/mol
O = 6 * 16.00 g/mol
Adding all the masses together:
Molar mass of C3H5(NO3)2 ≈ 3(12.01) + 5(1.008) + 2(14.01) + 6(16.00) g/mol
≈ 227.13 g/mol
Step 6: Convert moles of C3H5(NO3)2 to grams:
Mass = moles * molar mass
Mass = 0.9917 mol * 227.13 g/mol
Mass ≈ 225.42 g
Therefore, the minimum mass of C3H5(NO3)2 needed to be added to the dynamite at 25 °C is approximately 225.42 grams.
It's important to note that this calculation assumes ideal conditions and a complete reaction. In practice, other factors may also affect the explosive properties and the overall efficiency of the dynamite.