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mr_godi [17]
3 years ago
5

Initial mass of triphenyl methanol in g = 0.220g Initial volume of 33% HBr solution in ml = 0.60 ml Find mas of triphenyl bromid

e in g = 240 g
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

0.792g of triphenyl bromide are produced.

Explanation:

The reaction of triphenyl methanol with HBr is:

triphenyl methanol + HBr → Triphenyl bromide.

<em>Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.</em>

<em />

To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first <em>limiting reactant</em>:

Moles triphenyl methanol (Molar mass: 260.33g/mol) =

0.220g × (1mol / 260.33g) = <em>8.45x10⁻³ moles Triphenyl methanol</em>

Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =

0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = <em>2.45x10⁻³ moles HBr</em>

<em />

As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is <em>limiting reactant.</em>

<em />

As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = <em>2.45x10⁻³ moles</em>

As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:

2.45x10⁻³ moles × (323.2g / mol) =

<h3>0.792g of triphenyl bromide are produced.</h3>
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