How many gramsSO2 can be formed from 20 g S and 15.0 g of O2? ?

Question

Please help, I'm so confused. If you could actually answer it that would be great, or just walk me through it so I can answer it. Thanks

Answer 1

1)Write the equation for the reaction.

    S (s) + O2(g) ------> SO2 (g)

2)Determine the number of moles of everything present (ie) S and O2.

  Use the formula:

                   moles = mass/Mr

3) moles S (s) = 20/32

                    =  0.625

   moles O2(g) = 15/32

                    =  0.469

4) Now from here you must determine the limiting reagent (ie)the substance that will limit the amount of product formed due to its amount.

Note that there are fewer moles of oxygen than sulphur(S) and that they react in a 1:1 ratio.So when the 0.469 moles of O2(g) are used up the reaction will no longer form any SO2 (g) as there is no oxygen to react.Therefore this is the limiting reagent.

5)Now using the number of moles of O2(g) determine the  number of moles of SO2(g) that will be formed.

  1 mole O2(g)  : 1 mole SO2 (g)

0.469 moles O2 (g) : 0.469 moles So2(g).

6)Now;

     1 mole SO2  = 64 g

   0.469 moles SO2 = 64 * 0.469 =30.0 g

Good Luck!

Answer 2

S + O2 >> SO2

moles S = 20 g / 32.066 g/mol =0.624

Moles O2 = 15.0 g / 32 g/mol =0.469

O2 is the limiting reactant so we would get 0.469 moles of SO2

Mass SO2 = 0.469 mol x 64.066 g/mol = 30.0 g

Answer 3

-the problem asks you about making a compound from two different amounts of two reactants.  this should automatically clue you into thinking that this is a limited reactant question.

step1- set up a balanced equation for the problem

S + O2 -> SO2

luckily, this reaction is pretty straightforward.

so this equation means that 1 mole S + 1 mole O2 gives you 1 mole of SO2

step2- convert the grams of reactants to their amount in moles

20gS/1 x 1mol/32gS = 20/32 mols of S

15gO2/1 x 1mol/32gO2 = 15/32 mols of O2

step3- determining the limited reactant

since 1 mole S and 1 mole O2 makes 1 mole SO2, O2 will be the limited reactant because it will all be used up, while there is still some moles of S left because 20/32 > 15/32.

step4- convert number of moles of limited reactant to mols of product

so based on the balanced equation from above, from 15/32 moles of limited reactant, we get 15/32 moles of product

step5- convert moles of product back to grams of product (SO2)

15/32 moles SO2 x 64gSO2/1 mole SO2 = 30 grams of SO2

Answer 4

S + O2 --------> SO2

  Molecular mass of S =32 g/mol

  Molecular mass of O2=32 g/mol

Molecular mass of  SO2=32+32=64 g/mol

no. of moles of S = amount in gram/molecular mass

                          =20/32

                          =0.625mol

no.of moles of O2= 15/32

                          =0.468mol

1 mol of S react with 1 mol of O2 to give 1mol of SO2

0.625 mol of S will react with 0.625 mol of O2

but only 0.468mol is present

therefore amount of product formed will depend on O2

therefore 0.468mol of SO2 is formed

mass of SO2 = 0.468*64

                        =29.952

                          =30 g