Areaction was performed in which 1.400 g of camphor was reduced by an excess of sodium borohydride to make 1.061 g of isoborneol. calculate the theoretical yield and percent yield for this reaction.

The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %

Explanation:

Step 1: Data given

Camphor = C10H16O

sodium borohydride = NaBH4

Isoborneol = C10H18O

Mass of camphor = 1.400 grams

Mass of sodium borohydride = 1.061 grams

Molar mass of camphor = 152.23 g/mol

Molar mass of sodium borohydride = 37.83 g/mol

Molar mass of isoborneol = 154.25 g/mol

sodium borohydride is in excess, this means camphor is the limiting reactant

Step 2: The balanced equation

2C10H16O + NaBH4 → 2C10H18O + NaB

Step 3: Calculate moles of camphor

Moles camphor = mass camphor / molar mass camphor

Moles camphor = 1.400 grams / 152.23 g/mol

Moles camphor = 0.00920 moles

Step 4: Calculate moles of isoborneol

For 2 moles camphor we need 1 mol of NaBH4 to produce 2 moles of isoborneol

For 0.00920 moles camphor we'll have 0.00920 moles isoborneol

Step 5: Calculate mass isoborneol

Mass isoborneol = moles isoborneol * molar mass isoborneol

Mass isoborneol = 0.00920 * 154.25 g/mol

Mass isoborneol =  1.4191 grams (= theoretical yield)

% yield = (actual yield / theoretical yield) *100%

% yield = (1.061 / 1.4191) *100 %

% yield = 74.77 %

The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %

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%

lol idk man i guess u gotta ask bill nye the science guy