Chemical because the propane has stored chemical energy which is being released and a by-product of that propane is heat or themal energy.
Answer: 72 grams of
are needed to completely burn 19.7 g ![C_3H_8(g)](https://tex.z-dn.net/?f=C_3H_8%28g%29)
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number
of particles.
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Putting in the values we get:
![\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B19.7g%7D%7B44g%2Fmol%7D%3D0.45moles)
![C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)](https://tex.z-dn.net/?f=C_3H_8%28g%29%2B5O_2%28g%29%5Crightarrow%203CO_2%28g%29%2B4H_2O%28g%29)
According to stoichiometry:
1 mole of
requires 5 moles of oxygen
0.45 moles of
require=
moles of oxygen
Mass of ![O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g](https://tex.z-dn.net/?f=O_2%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D2.25%5Ctimes%2032%3D72g)
72 grams of
are needed to completely burn 19.7 g ![C_3H_8(g)](https://tex.z-dn.net/?f=C_3H_8%28g%29)
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