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.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.