Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide according to the following chemical equation : 2 S(s) + 3 O2(g) + 4 NaOH(aq) → 2 Na2SO4(aq) + 2 H2O(l)

Sodium hydroxide NaOH will be the limiting reagent.

Explanation:

By stocheimetry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

On the other hand, you know the following masses of each element that make up the compounds:

Therefore, the molar mass of each compound is:

Therefore, by stoichiometry of the reaction, the following amounts of the compounds react by mass or are produced:

Now, you apply a simple rule of three as follows: if 64 grams of S reacts with stoichiometry with 96 grams of O₂, 2 grams of S with how much mass of O₂ will react?

mass of O₂= 3 grams

You have enough amount of O₂ since it needs 3 grams and has 3 grams. The oxygen gas will not be the limiting reagent.

Now, you apply the following rule of three: if 64 grams of S reacts with stoichiometry with 160 grams of NaOH, 2 grams of S with how much mass of NaOH will react?

mass of NaOH= 5 grams

In this case you need 5 grams of NaOH, but it only has 4 grams. This indicates that you do not have enough, which means that sodium hydroxide will be the limiting reagent.

it's filtration, but this is an assumption so my apologies if i got this

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