Methane (CH4) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. (a) Write the balanced chemical equation for the reaction of methane and sulfur. (Use the lowest possible coefficients. Include states of matter under SATP conditions in your answer.)(b) Calculate the theoretical yield of carbon disulfide when g of methane is reacted with an equal mass of sulfur.

(a).  The balanced chemical reaction will be:

(b). 120 grams of sulfur will produce 71.25 grams of carbon disulfide.

The heating of methane in the presence of sulfur can be given by:

(a). The lowest possible coefficients have been used to balance the number of elements on both sides i.e. reactant and the product.

The state of the elements has also been mentioned in the chemical equation. The state of methane is gas, sulfur is solid, carbon disulfide is liquid and hydrogen disulfide is gas.

Thus the balanced chemical reaction will be:

(b). When 120 grams of sulfur reacts, the yield of carbon disulfide will be:

From the balanced equation, 4 moles of sulfur will give 1 mole of carbon disulfide.

4 moles of Sulfur = mole molecular weight

4 moles of Sulfur = 4 32

4 moles of Sulfur = 128 grams.

1 mole of carbon disulfide = 76 grams.

So, 128 grams of sulfur produces, 76 grams of carbon disulfide.

Thus, 120 grams of sulfur will produce = grams Carbon disulfide

120 grams of sulfur will produce = 71.25 grams of Carbon disulfide.

Thus, 120 grams of sulfur will produce 71.25 grams of carbon disulfide.

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71.25 g

Explanation:

Methane + Sulfur = Carbon Disulfide + Hydrogen Sulfide

The equation is given as;

CH4 + S --> CS2 + H2S

Upon balancing, we have;

CH4(g) + 4S(s) → CS2(l) + 2H2S(g)

Theoretical yield of carbon disulphide;

Equal masses of 120g

The limiting reactant is sulphur It determines how much of the product is formed.

From the equation;

4 mol of S produces 1 mol of CS2

upon converting to masses using; Mass = number of mol * Molar mass

4 mol of S = 4 * 32

1 mol of CS2 = 76g

128g of S produces 76g of C2, 120g would produce ?

128 = 76

120 = x

x = 120 * 76 / 128

x = 71.25g

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-