Consider the following balanced equation for the following reaction:
15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if the percent yield of CO2(g) is 83.0% and the theoretical yield of CO2(g) is 1.30 moles.

The amount of carbon dioxide produced in the reaction is 47.48 grams

Explanation:

For the given chemical equation:

To calculate the actual yield of carbon dioxide, we use the equation:

Percentage yield of carbon dioxide = 83 %

Theoretical yield of carbon dioxide = 1.30 moles

Putting values in above equation, we get:

To calculate the mass for given number of moles, we use the equation:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.079 moles

Putting values in above equation, we get:

Hence, the amount of carbon dioxide produced in the reaction is 47.48 grams

the answer for sodium is 0.186

the answer for rubidium is 0.251

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml