Cigarette lighters use butane, or C4H10, as their fuel. If 120.0 grams of butane burn with only 55.0 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant?

Volume CO₂ produced  := 33.85 L

Given

120.0 grams of butane

55.0 liters of oxygen

Required

Carbon dioxide produced

Solution

Reaction(combustion of Butane) :

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O.

mol of Butane (MW=58 g/mol) :

mol = mass : MW

mol = 120 : 58

mol = 2.069

At STP(1 atm, 273 K) , 1 mol gas=22.4 L, so for 55 L Oxygen :

mol O₂= 55 : 22.4 =2.455

Mol ratio of Butane and Oxygen = mol : reaction coefficient :

C₄H₁₀ : O₂ = 2.069/2 : 2.455/13 = 1.0345 : 0.189

Limiting reactant ⇒ O₂ (smaller ratio)

So mol CO₂ based on limiting reactant(O₂)

From equation, mol CO₂ : O₂ = 8 : 13, so mol CO₂ :

= 8/13 x mol O₂

= 8/13 x 2.455

= 1.511

Volume CO₂ at STP :

= 1.511 x 22.4 L

= 33.85 L

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms