Balance the following chemical equations using the smallest whole numbers (NO FR

Question

Balance the following chemical equations using the smallest whole numbers (NO FRACmoNS% Use numbers only not words. For example, write "3', not three. IF THE COEFFICIENT is 1, DO NOT LEAE IT BLANK!!: C3H6 (g) NH3 (g) O2 (g) (go H20 (g) For the following reaction: Before 15.0 g of H2S reacts with 10.00 g of SO2, the initial amount of moles of H2S are initial amount of moles of 2 are mol. (USE PROPER SIG FIGS FOR EACHn How many moles of H20 will be produced? mol How many grams of S will be produced? mol and the

Explanation / Answer

Answer –

We are given , reaction and we need to balance it to smallest whole number –

   C3H6(g) + NH3(g) + O2(g) ------> C3H3N(g) + H2O(g)

Now we need to balance each element at one time

    C3H6(g) + NH3(g) + O2(g) ------> C3H3N(g) + H2O(g)

     C = 3                                                C = 3

     H = 9                                                 H = 5

     N = 1                                                N = 1

     O = 2                                               O = 1

Now in this reaction there is already C and N balanced, now we need to balance the H element- In the reactant side there are 9 H and in the product side there is 5 only, so we need number even number for H

       C3H6(g) + 2 NH3(g) + O2(g) ------> C3H3N(g) + H2O(g)

     C = 3                                                C = 3

     H = 12                                               H = 7

     N = 2                                                N = 1

     O = 2                                               O = 1

Now we need to balance the N – in the reactant side there are 2 N and in product there is 1 N

So we need to put 2 in front of C3H3N

   C3H6(g) + 2 NH3(g) + O2(g) ------> 2 C3H3N(g) + H2O(g)

     C = 3                                                C = 6

     H = 12                                               H = 8

     N = 2                                                N = 2

     O = 2                                               O = 1

Now we need to balance the C – in the reactant side there are 3 C and in product there are 6 C

So we need to put 2 in front of C3H6

2 C3H6(g) + 2 NH3(g) + O2(g) ------> 2 C3H3N(g) + H2O(g)

     C = 6                                                C = 6

     H = 18                                               H = 8

     N = 2                                                N = 2

     O = 2                                               O = 1

Now we need to balance the H – in the reactant side there are 18 H and in product there are 8 H ,So we need to put 6 in front of H2O

2 C3H6(g) + 2 NH3(g) + O2(g) ------> 2 C3H3N(g) + 6H2O(g)

     C = 6                                                C = 6

     H = 18                                               H = 8

     N = 2                                                N = 2

     O = 2                                               O = 6

Now we need to balance the O – in the reactant side there are 2 O and in product there are 6 O, So we need to put 3 in front of O2

2 C3H6(g) + 2 NH3(g) + 3O2(g) ------> 2 C3H3N(g) + 6H2O(g)

     C = 6                                                C = 6

     H = 18                                               H = 8

     N = 2                                                N = 2

     O = 6                                               O = 6

Now this reaction get balanced

So balanced reaction is –

2 C3H6(g) + 2 NH3(g) + 3O2(g) ------> 2 C3H3N(g) + 6H2O(g)

Now in second question we are given,

2H2S + SO2 ---> 3S + 2 H2O

Mass of H2S = 15.0 g , mass of SO2 = 10.00 g

Moles of H2S = 15.0 g / 34.0818 g.mol-1

                       = 0.440 moles of H2S

Moles of SO2 = 10.00 g / 64.064 g.mol-1

                        = 0.156 moles of SO2

Limiting reactant –

We need to calculate moles of S from both the reactant

Moles of S from H2S

From the balanced equation –

2 moles of H2S = 3 moles of S

So, 0.440 moles of H2S = ?

= 0.440 moles of H2S * 3 moles of S / 2 moles of H2S

= 0.660 moles of S

Moles of S from SO2

From the balanced equation –

1 moles of SO2 = 3 moles of S

So, 0.156 moles of SO2 = ?

= 0.156 moles of SO2* 3 moles of S / 1 moles of SO2

= 0.468 moles of S

So moles of S getting lowest from the SO2, so SO2 is the limiting reactant.

Mole of H2O from the limiting reactant

1 moles of SO2 = 2 moles of H2O

So, 0.156 moles of SO2 = ?

= 0.156 moles of SO2* 2 moles of H2O / 1 moles of SO2

= 0.313 moles of H2O

Mass of S = 0.468 moles * 32.066 g/mol

                  = 15.0 g of S

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