MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
In this question, you are given the NaOH volume but asked for concentration.
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:
OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3
The best answer is (3)
In these kind of reactions, there is a transfer of electrons from one reactant to another. electrons are lost from one substance and gained by another.
Oxidation is loss of electrons from a substance, and Reduction is gain of electrons by a substance.
These two processes cannot occur without the other. If there is a reduction there must be an oxidation reaction and vice versa. The reactions usually occur simultaneously.
For example, table salt is formed by a redox reaction. Sodium is oxidized i.e. loses an electron (and becomes positively charged) while chlorine gas is reduced i.e. gains the electron (and become negatively charged). The result is formation of sodium chloride.
False, many chemicals can be toxic to the drainage systems, and can make their way into the ocean harming sea life.
Answer:
3) ester
Explanation:
Esterification is the process in which alkanol and alkanoic acids reacts in the presence of a catalyst and heat. The product is usually an ester(alkanoate) and water.
For the reaction between ethanoic acid and 1-butanol, the product is butylethanoate and water as shown below:
CH₃COOH + C₄H₈OH → CH₃COOC₄H₈ + H₂O
