Aluminum reacts with oxygen to yield aluminum oxide. if 5.0 g of al reacts with 4.45 g of o2, what is the empirical formula of aluminum oxide? (type your answer using the format co2 for co2.)

Let us assume that the given values of each element completely react with each other. To solve for the empirical formula, first we have to calculate for the number of moles of each element.

Molar mass Al = 27 g / mol

Molar mass Oxygen = 16 g / mol

Therefore the number of moles each is:

moles Al = 5 / 27 = 0.185 mol

moles O = 4.45 / 16 = 0.278 mol

Next we divide each moles with the smallest mol value, in this case 0.185 mol, therefore divide the two with 0.185 mol:

moles Al = 0.185 / 0.185 = 1

moles O = 0.278/0.185 = 1.5

The empirical formula must be in whole numbers, however moles O is still in decimal. Therefore multiply the two by 2:

moles Al = 1 * 2 = 2

moles O = 1.5 * 2 = 3

Therefore the empirical formula is:

Al2O3

dynamic equilibrium is the answer.

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