The reaction of Butanoic acid and excess ethanol occurs as shown in the equation;
CH₃CH₂CH₂COOH + CH₃CH₂OH = CH₃CH₂CH₂COOCH₂CH₃ + H₂O
1 mole of butanoic acid contains 88g/mol
Thus, 7.35 g of butanoic acid contains;
= 7.35/88
= 0.0835 moles
The mole ratio of butanoic acid and ethylbutyrate is 1:1
Thus moles of ethyl butyrate produced is 0.0835 moles
But 1 mole of ethyl butyrate contains 116 g/mol
Hence, the mass of ethyl butyrate will be
=0.0835 × 116
=9.686 g
Cl has 7 valence electrons, it needs 8 to be stable, according to the octet rule, so Cl gains stability by gaining 1 electron to reach 8 electrons
Answer:- Yes, strontum bromide and potassium sulfate gives a precipitate of strontium sulfate.
Explanations:- As per the solubility rules, all compounds of alkali metals are soluble.
Sulfate of most of the alkaline earth metals like Ca, Ba and Sr are insoluble.
A double displacement reaction takes place when strontium bromide and potassium sulfate are mixed and a precipitate of strontium sulfate is formed:
Note: (aq) stands for aqueous and (s) stands for solid and here it's precipitate.
Answer:
The limiting reagent is the SF₄
Explanation:
In order to determine the limiting reagent we convert the mass of the reactants to moles, and then, we work with stoichiometry.
4.687 g / 108.06 g/mol = 0.0433 moles of SF₄
6.281 g / 333.8 g/mol = 0.0188 moles of I₂O₅
The reaction is: 5SF₄ + 2I₂O₅ = 4IF₅ + 5SO₂
Ratio is 5:2. 5 moles of fluoride react with 2 moles of pentoxide
Then, 0.0433 moles of fluoride will react with (0.0433 . 2) / 5 = 0.0173 moles
We have 0.0188 moles of I₂O₅ and we need 0.0173 so there are some moles of pentoxide that remains after the reaction. In conclussion, the limiting reagent is the SF₄. We verify:
2 moles of pentoxide react with 5 moles of SF₄
Therefore, 0.0188 moles of I₂O₅ will react with (0.0188 . 5) / 2 = 0.0470 moles.
As we have 0.0433 moles of SF₄, we do not have enough moles.