Ethanol has the chemical formula C2 H6 O . A sample of ethanol contains 164grams of C. How many molecules are in the sample? (4.11E24 molecules C2 H6 O)

We have 8.22 1 C atoms

We have 4.11 1 ethanol molecules

Explanation:

Step 1: Data given

Molecular formula of ethanol =

Mass of C = 164 grams

Molar mass C = 12.01 g/mol

Step 2: Calculate moles C

Moles C = mass C / molar mass C

Moles C = 164 grams / 12.01 g/mol

Moles C = 13.66 moles

Step 3: Calculate molecules of C

Atoms C = moles C number of Avogadro

Atoms C = 13.66 moles 6.02 1 molecules/ mol

Atoms C = 8.22 1 C atoms

Step 4: Calculate molecules in the sample

For 1 mol we have 2 moles C

For 13.66 moles C we have 13.66/ 2 = 6.83 moles

Molecules = 6.83 moles 6.02 1 = 4.11 1 molecules.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%