Using this balanced equation: 2 naoh + h2so4 —> h2o + na2so4

how many grams of sodium sulfate will be formed if you start with 150 grams of
sodium hydroxide and you have an excess of sulfuric acid?

!

266.325 g

We are given the balanced equation;

2NaOH + H₂SO₄ → H₂O + Na₂SO₄

We are required to determine the mass of Na₂SO₄ that will be formed.

Moles = Mass ÷ molar mass

Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 150 g ÷ 40 g/mol

                          = 3.75 moles

Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2

                                      = 3.75 moles ÷ 2

                                     = 1.875 moles

we know that;

Mass = Moles × Molar mass

Molar mass of Na₂SO₄ is 142.04 g/mol

Therefore;

Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol

                           = 266.325 g

Thus, the mass of sodium sulfate formed 266.325 g

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