How many grams of NaN3 are needed to produce 20.0 L of N2(g) at 30.0 oC and 776 mmHg ?
well, 2 moles of NaN3 will give 3 moles of N2
soconverting 20 l at above, to STP...
moles=20/22.4 *(273/303 )=.80 moles
so we want 2/3 * .8 moles naN3, or grams= 1.6/3 * 65g=34.7 grams
To determine the number of grams of NaN3 needed to produce a specific amount of N2 gas, we need to use the Ideal Gas Law and the balanced chemical equation for the reaction.
First, let's write the balanced chemical equation for the reaction of NaN3 decomposing to produce N2 gas:
2NaN3(s) -> 2Na(s) + 3N2(g)
From the equation, we can see that 2 moles of NaN3 produce 3 moles of N2 gas.
Next, let's use the Ideal Gas Law equation to find the number of moles of N2 gas:
PV = nRT
Where:
P = pressure of the gas (776 mmHg)
V = volume of the gas (20.0 L)
n = number of moles of the gas (unknown)
R = ideal gas constant (0.0821 L * atm / (mol * K))
T = temperature in Kelvin (30.0 oC + 273.15 = 303.15 K)
Rearranging the equation to solve for n:
n = PV / RT
Substituting the given values:
n = (776 mmHg * 20.0 L) / (0.0821 L * atm / (mol * K) * 303.15 K)
Note: 1 atm = 760 mmHg
Now, we can calculate the number of moles of N2 gas.