Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements.
2 NaN3(s) 2 Na(s) + 3 N2(g)
(b) How many grams of NaN3 are required to form 16.0 g of nitrogen gas?
(c) How many grams of NaN3 are required to produce 18.0 ft3 of nitrogen gas if the gas has a density of 1.25 g/L?
2 NaN3(s) 2 Na(s) + 3 N2(g)
(b) How many grams of NaN3 are required to form 16.0 g of nitrogen gas?
(c) How many grams of NaN3 are required to produce 18.0 ft3 of nitrogen gas if the gas has a density of 1.25 g/L?
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2 NaN3(s) ---> 2 Na(s) + 3 N2(g)
b) We need to first calculate how many moles of Nitrogen gas are in 16 grams of... We can do that by dividing the 16 grams by the molar mass of Nitrogen gas.
16.0 g / 28.0 g/mol = 0.571 mol N2
Now, we ratio of moles of NaN3 to N2 is 2:3 from the balanced reaction. We can use that to calculate how many moles of NaN3 there are.
(2 mol NaN3 / 3 mol N2) (0.571 mol N2) = 0.381 mol NaN3
Now we just multiply the moles of NaN3 by its molar mass to get the required grams of...
0.381 mol x 65.0 g/mol = 24.8 g NaN3
c) This is the doing part b again, but with some density calculations. First, you have to convert the cubic feet to liters.
1 ft^3 ---> 28.32 L
18.0 ft^3 ---> 509.7 L
Now, we have to use the volume in the density formula to get the mass of N2 gas.
D = m/V
1.25 g/L = m / 509.7 L
m = 1.25 g/L x 509.7 L
m = 637. g N2
Now, we know that 16 g of N2 gas requires 24.8 g of NaN3. We need to know how many grams of NaN3 are required for 637 grams of N2. We can just set up a proportion to solve this.
(16 / 24.8) = (637 / grams NaN3)
Grams of NaN3 x 16 = 24.8 x 637
Grams of NaN3 = (24.8 x 637) / 16
Grams of NaN3 = 987. g NaN3
Hope I helped :)
b) We need to first calculate how many moles of Nitrogen gas are in 16 grams of... We can do that by dividing the 16 grams by the molar mass of Nitrogen gas.
16.0 g / 28.0 g/mol = 0.571 mol N2
Now, we ratio of moles of NaN3 to N2 is 2:3 from the balanced reaction. We can use that to calculate how many moles of NaN3 there are.
(2 mol NaN3 / 3 mol N2) (0.571 mol N2) = 0.381 mol NaN3
Now we just multiply the moles of NaN3 by its molar mass to get the required grams of...
0.381 mol x 65.0 g/mol = 24.8 g NaN3
c) This is the doing part b again, but with some density calculations. First, you have to convert the cubic feet to liters.
1 ft^3 ---> 28.32 L
18.0 ft^3 ---> 509.7 L
Now, we have to use the volume in the density formula to get the mass of N2 gas.
D = m/V
1.25 g/L = m / 509.7 L
m = 1.25 g/L x 509.7 L
m = 637. g N2
Now, we know that 16 g of N2 gas requires 24.8 g of NaN3. We need to know how many grams of NaN3 are required for 637 grams of N2. We can just set up a proportion to solve this.
(16 / 24.8) = (637 / grams NaN3)
Grams of NaN3 x 16 = 24.8 x 637
Grams of NaN3 = (24.8 x 637) / 16
Grams of NaN3 = 987. g NaN3
Hope I helped :)
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(b) 16 g N2 ( 1 mol N2 / 28 g N2) = 0.571 mol N2
0.571 mol N2 X 2 mol NaN3 / 3 mol N2) = 0.381 mol NaN3
0.381 mol NaN3 X 65.0 g/mol = 24.8 g NaN3
(c) I'll tell you the steps, but you'll have to do the calculations:
1. Convert 18 ft^3 into Liters
2. multiply Liters by the density 1.25 g/L to get grams of N2 produced
3. Then, solve the problem as you did in (b), converting grams of N2 into moles of N2, using the equation to relate moles of N2 to moles of NaN3, and finally using the molar mass of NaN3 to calculate grams NaN3.
0.571 mol N2 X 2 mol NaN3 / 3 mol N2) = 0.381 mol NaN3
0.381 mol NaN3 X 65.0 g/mol = 24.8 g NaN3
(c) I'll tell you the steps, but you'll have to do the calculations:
1. Convert 18 ft^3 into Liters
2. multiply Liters by the density 1.25 g/L to get grams of N2 produced
3. Then, solve the problem as you did in (b), converting grams of N2 into moles of N2, using the equation to relate moles of N2 to moles of NaN3, and finally using the molar mass of NaN3 to calculate grams NaN3.
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It's AMAZINGGood Luck, cheers
http://newscope.info/239637/nitrogen-gas
http://newscope.info/239637/nitrogen-gas