4C₃H₅(NO₃)₃
------> 12CO₂
+ 6N₂ + 10H₂O + O₂
mol of CO₂ = 
= 
mol ratio of CO₂ : C₃H₅(NO₃)₃
12 : 4
∴ if mole of CO₂ = 0.568 mol
then " " C₃H₅(NO₃)₃ = 
= 0.189 mol
∴ mass of nitroglycerin = mole * Mr
= 0.189 mol * 227.0995 g / mol
= 43.00 g
Answer:
c
Explanation:
was c the correct answer or not
Answer:
1. 3.83 L
2. 0.368 mole
Explanation:
1. Determination of the volume
Pressure (P) = 3.21 atm
Temperature (T) = 202 K
Number of mole (n) = 0.741 mole
Gas constant (R) = 0.0821 L.atm/molK
Volume (V) =?
The volume can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
3.21 × V = 0.741 × 0.0821 × 202
3.21 × V = 12.3497283
Divide both side by 3.21
V = 12.2888922 / 3.21
V = 3.83 L
Thus, the volume of the gas is 3.83 L
2. Determination of the number of mole.
Pressure (P) = 2.50 atm
Temperature (T) = 215 K
Volume (V) = 2.60 L
Gas constant (R) = 0.0821 L.atm/molK
Number of mole (n) =?
The number of mole can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
2.50 × 2.60 = n × 0.0821 × 215
6.5 = n × 17.6515
Divide both side by 17.6515
n = 6.5 / 17.6515
n = 0.368 mole
Thus, the number of mole of the gas is 0.368 mole.
6 Li + N2 = 2 Li3N
6 moles Li ------------ 2 moles Li3N
0.24 moles Li ------- moles Li3N
moles Li3N = 0.24 *2 / 6
moles Li3N = 0.48 / 6 => 0.08 moles
Answer B
