# Question #7c266

##### 1 Answer

It will react with 191.7 gram

#### Explanation:

To do a calculation on a reaction, we first determine the mol ratio in which the reactants reacts and the products are formed. This ratio is already given in the equation by the coefficients.

#color(red)(color(white)(aaa)1 color(white)(aaaaaaa): color(white)(aaa)3 color(white)(aaa): color(white)(aaaa)2 color(white)(aaaa): color(white)(aaa)3)#

#C_2H_5OH (g) + 3 O_2 (g) -> 2 CO_2 (g) + 3H_2O (g)#

To find out the mass of both the products, we use the law of conservation of mass, that states that mass cannot be created nor destroyed.

We have given that we burn 92 gram of

The molar mass of

2 times a C-atom

#=> 2xx12.01 =24.02# g/mol

6 times a H-atom#=>6xx1.008=6.048# g/mol

1 time an O-atom#=>1xx16.00=16.00# g/mol

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)color(white)(aaaaaaaaaaa)/color(white)(a) +#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)46.068# gram/mol

*Note: We also find the molar mass for the 3 other compounds, but the calculations aren't shown here since they are all similar.*

Now we can use the mol ratio to calculate the number of moles that is reacted or is produced with 1.997 mol

**calculations**

#C_2H_5OH (g) + 3 O_2 (g) #

#color(red)(color(white)(aaa)1 color(white)(aaaaaaa): color(white)(aaa)3 )#

#color(white)(aaaaaaaaaaa)cancel"\"#

#color(white)(aaa)1.997 color(white)(aaaaaaa) color(blue)(5.991 " mol") #

5.991

**calculations**

#C_2H_5OH (g) -> 2 CO_2 (g) #

#color(red)(color(white)(aaa)1 color(white)(aaaaaaa): color(white)(aaa)2 )#

#color(white)(aaaaaaaaaaa)cancel"\"#

#color(white)(aaa)1.997 color(white)(aaaaaaa) color(blue)(3.994 " mol") #

3.994

**calculations**

#C_2H_5OH (g) -> 3H_2O (g) #

#color(red)(color(white)(aaa)1 color(white)(aaaaaaa): color(white)(aaa)3 )#

#color(white)(aaaaaaaaaaa)cancel"\"#

#color(white)(aaa)1.997 color(white)(aaaaaaa) color(blue)(5.991 " mol") #

5.991

The 3 reaction schemes showed above are just to visualise what we are doing with the mol ratio. These scheme's do not represent a chemical reaction. The calculations are more easily done when writing everything under the total equation (the one you provided) instead of first making the separate schemes, but this is just to show how it works..