One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is below. fe2o3 + 3 co → 2 fe + 3 co2 suppose that 1.92 ✕ 103 kg of fe are obtained from a 5.13 ✕ 103 kg sample of fe2o3. assuming that the reaction goes to completion, what is the percent purity of fe2o3 in the original sample?

The answer to your question is: 53.46 % pure

Explanation:

data

Fe = 1.92 ✕ 103 kg   produced    = 1920 kg

Fe2O3 = 5.13 ✕ 103 kg    sample   = 5130 kg

MW Fe2O3 = (56x2)+(16x3) = 160 kg

% of purity = ?

                   Fe2O3 + 3 CO → 2 Fe + 3 CO2

Convert mass to moles

Fe

      56 kg   1 mol

  1920 kg   x moles       x = 34.28 moles

From the reaction

   1 mol of Fe2O3  2 moles of Fe

      x moles of Fe2O3 34.28 moles

  x = 34.28/2 = 17.14 moles of Fe2O3

       160 kg of FE2O3  1 mol

             x kg of Fe2O3  17.14 moles

x = 17.14 x 160/1 = 2742,4 kg of Fe2O3  It's supposed to be the amount of Fe if it was 100% pure.

                             2742.4 kg of Fe2O3  100%

                              5130 kg of Fe2O3   x

x = (2742.4x100)/5130 = 53.46 pure

ill try to . no promises tho.

explanation:

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