11) if 352 moles of butane (ch) is mixed with 8.47 moles of oxygen gas, how many grams of water will be produced? 2c, h,0 + 130, 800, + 10 h,0 (balanced)

117 g  

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:       63.55      32.00                    18.02

             2C₄H₁₀  +  13O₂  ⟶ 8CO₂ + 10H₂O

n/mol:     352         8.47

2. Calculate the amount of water that each reactant can produce

(a) From C₄H₁₀

Moles of H₂O = 352 mol C₄H₁₀ × (10 mol H₂O/2 mol C₄H₁₀) = 704 mol H₂O

(b) From O₂:

Moles of H₂O = 8.47 mol O₂ × (10 mol H₂O/13 mol O₂) = 6.515 mol H₂O

3. Identify the limiting reactant

The limiting reactant is O₂, because it gives fewer moles of H₂O.

4. Calculate the mass of water

Mass of H₂ = 5.515 mol H₂O × (18.02 g H₂)/(1 mol H₂O) = 117 g H₂O

The reaction produces 117 g H₂O.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72

it takes on an electronical energy