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igomit [66]
3 years ago
13

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

ater . Suppose 2.94 g of sulfuric acid is mixed with 1.6 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
quester [9]3 years ago
4 0

Answer:

0.72g

Explanation:

Step 1:

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

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

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

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

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olya-2409 [2.1K]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

Mass=Density\times Volume

Volume of solution = Volume of HCl + Volume of Ba(OH)_2

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

Mass=1g/mL\times 93.1mL=93.1g

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat released = ?

m = mass = 93.1 g

C_p = specific heat capacity of water = 4.184J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 29.83^oC

Now put all the given value in the above formula, we get:

Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC

Q=1881.43J=1.88kJ        (1 kJ = 1000 J)

Now we have to calculate the moles of Ba(OH)_2 and HCl.

\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}

\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol

and,

\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}

\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

From the balanced reaction we conclude that

As, 1 mole of Ba(OH)_2 react with 2 mole of HCl

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and Ba(OH)_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

As, 1 mole of Ba(OH)_2 react to give 2 mole of H_2O

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of H_2O

Now we have to calculate the change in enthalpy of the reaction.

\Delta H_{rxn}=-\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

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Suppose that the root-mean-square velocity vrms of water molecules (molecular mass is equal to 18.0 g/mol) in a flame is found t
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Answer:

T=954.41\ K

Explanation:

The expression for the root mean square speed is:

C_{rms}=\sqrt {\dfrac {3RT}{M}}

R is Gas constant having value = 8.314 J / K mol  

M is the molar mass of gas

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Temperature = ?

C_{rms}=1150  m/s

1150=\sqrt{\frac{3\times 8.314\times T}{0.018}}

1322500=\frac{24.942T}{0.018}

T=954.41\ K

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Answer:

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