What mass of solid aluminum hydroxide can be produced when 50.0mL of 0.200M aluminum nitrate is added to 200.0mL of 0.100M potassium hydroxide

520 mg Al(OH)₃

Explanation:

The reaction that takes place is:

Now we calculate how many moles of each reagent were added, using their concentrations and volumes:

10 mmol of Al(NO₃)₂ would react completely with (10*3) 30 mmol of KOH, there are not as many KOH mmol, so KOH is the limiting reactant.

We calculate how many moles of Al(OH)₃ are produced, using the limiting reactant:

Finally we convert mmol of Al(OH)₃ to mg, using its molar mass:

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