Respuesta :
Answer:
0.264 g of [tex]CO_2[/tex] can be formed from 288 mg of [tex]O_2[/tex]
Explanation:
The balanced chemical equation is
[tex]2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O[/tex]
The conversions are
Mass in mg [tex]O_2[/tex] is converted to mass in g [tex]O_2[/tex]
Mass in g [tex]O_2[/tex] is converted to moles [tex]O_2[/tex] by dividing with molar mass
Moles [tex]O_2[/tex] is converted to moles [tex]CO_2[/tex] by using the mole ratio of [tex]O_2:CO_2[/tex] is 9 : 6
Moles [tex]CO_2[/tex] is converted to mass [tex]CO_2[/tex] by multiplying with molar mass [tex]CO_2[/tex]
mass in mg [tex]O_2[/tex] > mass in g [tex]O_2[/tex] >moles [tex]O_2[/tex] > moles [tex]CO_2[/tex] > mass [tex]CO_2[/tex]
[tex]288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}[/tex]
=0.264g (Answer)
