Which compound could serve as a reactant in a neutralization reaction?(1) NaCl (3) CH3OH(2) KOH (4) CH3CHO

Chemistry

2 answers:

NikAS [45]2 years ago

8 0

KOH could serve as a reactant in a neutralization reaction

Serjik [45]2 years ago

5 0

Answer: (2) $KOH$

Explanation: When an acid reacts with the base then it undergoes neutralization to form a salt and water.

An acid is a substance that ionizes in the water to give hydrogen ion and have pH less than 7.

$HX\rightarrow H^++X^-$

A base is a substance that ionizes in the water to give hydroxide ion and has pH more than 7.

$MOH\rightarrow M^++OH^-$

Thus $KOH$ being a base can be a reactant in neutralization reaction.

$NaCl$ being a salt can be a product in a neutralization reaction.

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$