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3 years ago

If 24.1 g of sodium hydroxide react with 22.0 g of hydrochloric acid to form 35.3g

Chemistry

1 answer:

disa [49]3 years ago

4 0

Answer:

10.85 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaOH + HCl —> NaCl + H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation.

These can be obtained as illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Mass of NaOH from the balanced equation = 1 × 40 = 40 g

Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O from the balanced equation = 1 × 18 = 18 g

Summary:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Finally, we shall determine the mass of water, H2O produced from the reaction as follow:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Therefore, 24.1 g of NaOH will react to produce = (24.1 × 18)/40 = 10.85 g of H2O.

Therefore, 10.85 g of H2O were produced from the reaction.

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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine if each addition would exceed the capacity of the b

likoan [24]

Answer:

no one additions exceed the capacity of the buffer

Explanation:

given

Volume buffer = 500.0 mL = 0.5 L

mol HNO₂ = 0.5 L × 0.100 mol/L = 0.05 mol HNO₂

mol NO₂⁻ = 0.5 L × 0.150 mol/L = 0.075 mol NO₂⁻

solution

we know when any base more than 0.05 (HNO2) than exceed buffer capacity

and when any base more than 0.075 (KNO2) than exceed buffer capacity

when we add 250 mg NaOH (0.250 g)

than molar mass NaOH =40 g/mol

and mol NaOH = 0.250 g ÷ 40g/mol

mol NaOH = 0.00625 mol

0.00625 mol NaOH will be neutralized by 0.00625 mol HNO₂

so it would not exceed the capacity of the buffer.

and

when we add 350 mg KOH (0.350 g)

than molar mass KOH =56.10 g

and mol KOH = 0.350 g ÷ 56.10 g/mol

mol KOH = 0.0062 mol

here also capacity of the buffer will not be exceeded

and

now we add 1.25 g HBr

than molar mass HBr = 80.91 g/mol

and mol HBr = 1.25 g ÷ 80.91 g/mol

mol HBr = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻

so the capacity will not be exceeded.

and

we add 1.35 g HI

molar mass HI = 127.91 g/mol

so mol HI = 1.35 g ÷ 127.91 g/mol

mol HI = 0.011 mol

capacity of the buffer will not be exceed

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3 years ago

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kondaur [170]

Answer:

zinc and lead or copper and tin

Explanation:

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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$