Asked by Nancy
2 NaN3(s) = 2 Na(l) + 3 N2(g)
How many grams of NaN3 are needed to produce 20L of N2(g) at 303K and 776mmHg?
How many grams of NaN3 are needed to produce 20L of N2(g) at 303K and 776mmHg?
Answers
Answered by
DrBob222
Use PV = nRT. Substitute and solve for n = number of mols.
Then use the coefficients in the balanced equation to convert mols N2 to mols NaN3.
Finally, g NaN3 = mols NaN3 x molar mass NaN3.
Then use the coefficients in the balanced equation to convert mols N2 to mols NaN3.
Finally, g NaN3 = mols NaN3 x molar mass NaN3.
Answered by
jack
In the chemical reaction used in automotive air-bag safety systems, nitrogen gas is produced by the decomposition of sodium azide as described in the reaction below.
NaN3(s) Na(l) + N2(g)
What mass of NaN3 must be decomposed to generate enough nitrogen gas at 25.0 oC and 0.980 atm to fill an airbag with a volume of 5.0 x 104 cm3
NaN3(s) Na(l) + N2(g)
What mass of NaN3 must be decomposed to generate enough nitrogen gas at 25.0 oC and 0.980 atm to fill an airbag with a volume of 5.0 x 104 cm3
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