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PSYCHO15rus [73]
3 years ago
6

How many milligrams of product can be produced from the complete Diels-Alder reaction of 180. mg of anthracene and 100. mg of ma

leic anhydride? Enter only the number to three significant figures.
Chemistry
1 answer:
ZanzabumX [31]3 years ago
3 0

Answer: It will be produced 276,3 mg of product

Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:

C14H10 + C4H2O3 → C18H12O3

The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.

1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.

As 1 mol of C18H12O3 equals 276,3 g, the mass produced is 276,3 mg.

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ompare your TLC analysis with the results of the column chromatography. More polar molecules move more slowly through silica tha
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Biphenyl will have a higher R value than the Methyl Orange.

Explanation:

Biphenyl is a aromatic hydrocarbon and it is a nonpolar molecule.

Methyl Orange is a organic compound with a -SO₃⁻Na⁺ polar functional group which will induce a high polarity in the compound.

You may find the chemical structures of both molecules in the attached picture.

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Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react to
NikAS [45]

0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

<h3>What is baking powder?</h3>
  • Baking powder is a dry chemical leavener composed of carbonate or bicarbonate and a weak acid.
  • The addition of a buffer, such as cornstarch, prevents the base and acid from reacting prematurely.
  • Baking powder is used in baked goods to increase volume and lighten the texture.

To find how many moles of CO₂ are produced from 1.00 g of baking powder:

The balanced equation is:

  • Ca(H₂PO₄)₂(s) + 2NaHCO₃(s) → 2CO₂(g) + 2H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)

On 3.50 g of baking power:

  • mCa(H₂PO₄)₂ = 0.35 × 3.50 = 1.225 g
  • mNaHCO₃ = 0.31 × 3.50 = 1.085 g

The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol.

So,

  • Ca(H₂PO₄)₂: 40 + 4 × 1 + 31 + 8 × 16 = 203 g/mol
  • NaHCO₃: 23 + 1 + 12 + 3 × 16 = 84 g/mol

The number of moles is the mass divided by molar mass, so:

  • nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol
  • nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting.

Testing for Ca(H₂PO₄)₂, the stoichiometry is:

  • 1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol

So, NaHCO₃ is in excess.

The stoichiometry calculus must be done with the limiting reactant, then:

  • 1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol of CO₂

Therefore, 0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

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The correct question is given below:

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 3.50 g of baking powder?

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