Respuesta :
Answer: 72 grams of [tex]O_2(g)[/tex] are needed to completely burn 19.7 g [tex]C_3H_8(g)[/tex]
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Putting in the values we get:
[tex]\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles[/tex]
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
According to stoichiometry:
1 mole of [tex]C_3H_8[/tex] requires 5 moles of oxygen
0.45 moles of [tex]C_3H_8[/tex] require= [tex]\frac{5}{1}\times 0.45=2.25[/tex] moles of oxygen
Mass of [tex]O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g[/tex]
72 grams of [tex]O_2(g)[/tex] are needed to completely burn 19.7 g [tex]C_3H_8(g)[/tex]
