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.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

i believe the answer is true.