Question

How many mL of C8H18 are needed to react with 0.0500 mol O2? 2 C8H18 (l) + 25 O2 (g) - 16 CO2(g) + 18 H20 (g) molar masses: C8H18 = 114.22g/mole H2O = 18.02 g/mole CO2 = 44.01 g/mole O2 = 32.00 g/mole and make sure the answer has significant figures

Answer 1

0.45688mL

Step-by-step explanation:

1) First you should look on the coefficients.

`2C_8H_(18(l))\ + 25O_2_((g))=\ 16CO_(2(g))+ 18H_2O_((g))`

2) Do the ratio between C8H18 and O2:

`C_8H_(18)\ \ \ \ \ \ \ \ \ O_2\\2 : \ \ \ \ \ \ \ \ \ \ \ \ \ 25\\X: \ \ \ \ \ \ \ \ \ \ \ 0.0500moles\\2X0.0500/25= 0.004\\X=0.004 \moles\ of\ C_8H_(18)`

3) grams= moles X molecular mass

`0.004*114.22= 0.45688g`

[1g=1mL] => 0.45688mL