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2 years ago

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Chemistry

1 answer:

Mariulka [41]2 years ago

5 0

Answer:

Amount of $H_{2}O$ produced will be half of its original value.

Explanation:

Balanced reaction: $HCl+NaOH\rightarrow NaCl+H_{2}O$

According to balanced equation, 1 mol of HCl reacts with 1 mol of NaOH to produce 1 mol of $H_{2}O$

If amount of reactants (NaOH and HCl) are halved then we can write-

0.5 mol of HCl reacts with 0.5 mol of NaOH to produce 0.5 mol of $H_{2}O$ .

So, it is evident that amount of $H_{2}O$ to be produced will be half if amount of reactants are halved.

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The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close

Natasha_Volkova [10]

Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log( $A^-/HA$ ) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log( $A^-/HA\\$ ) and so 10 = $A^-/HA$ or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

6 0

2 years ago

A 2135 cm3 sample of dry air has a pressure of 98.4 kpa at 127 degrees Celsius. What is the volume of the sample if the Temperat

kykrilka [37]

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2

2135 cm3 divided by 127 degrees celcius = x divided by 206

answer: 3460 cm3

7 0

2 years ago

Read 2 more answers

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solu

irinina [24]

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = $6.05\times 10^3 kg=6.05\times 10^6 g$

$1 kg = 10^3 g$

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = $\frac{95.0}{100}\times 6.05\times 10^6 g$

$=5,747,500 g$

Moles of sulfuric acid = $\frac{5,747,500 g}{98 g/mol}=58,647.96 mol$

$H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O$

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

$\frac{1}{1}\times 58,647.96 mol=58,647.96 mol$ of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

$106 g/mol\times 58,647.96 mol=6,216,683.76 g$

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

3 0

2 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.314 kg of water?

Mariulka [41]

Answer:

g NaCl = 424.623 g

Explanation:

C NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent

∴ solvent: H2O

∴ mass H2O = 2.314 Kg

mol NaCl:

⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl

∴ mm NaCl = 58.44 g/mol

⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl

5 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago