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Aqueous sulfuric acid h2so4 will react with solid sodium hydroxide naoh to produce aqueous sodium sulfate na2so4 and liquid water h2o . suppose 89.3 g of sulfuric acid is mixed with 96. g of sodium hydroxide. calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. be sure your answer has the correct number of significant digits.

Respuesta :

Edufirst
1) Balanced chemical equation

H2SO4 + 2NaOH ---> Na2 SO4 + 2H2O

=> 1 mol H2SO4 : 2 moles NaOH

2) Convert 89.3 g of H2SO4 and 96.0 g of NaOH to moles

Molar mass of H2SO4 = 98.1 g/mol

Molar mass of NaOH = 40.0 g/mol

moles = mass in grams / molar mass

moles H2SO4 = 89.3 g / 98.1 g/mol = 0.910 mol

moles NaOH = 96.0 g / 40.0 g/mol = 2.40 mol

3) Theoretical molar ratio = 2 moles NaOH / 1 mol H2SO4

So, all the 0.91 mol of H2SO4 will be consumed along with 1.820 (2*0.91) moles of NaOH, and 0.580 moles (2.40 - 1.82) of NaOH will be left over by the chemical reaction.

4) Convert 0.580 moles NaOH to mass

0.580 moles * 40.0 g/mol = 23.2 g of NaOH will be left over

ankitprmr2

Answer:

[tex]\bold{0.580 \;\rm{moles} \times 40.0\;\rm{ g/mol}} = 23.2\; g\; of\; NaOH[/tex] will be left.

Explanation:

Given:

Aqueous sulfuric acid [tex]H_2SO_4[/tex] will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate [tex]Na_2SO_4[/tex] and liquid water [tex]H_2O[/tex].

Now, balance the molecules of the left part of the equation with the right part of the equation.  

[tex]H_2SO_4 + 2NaOH \rightarrow Na_2 SO_4 + 2H_2O[/tex]

Now, from the above-balanced equation, it is clear that  1-mole sulphuric acid equated with two moles of sodium hydroxide to balance the above equation.

Now,

The weight of sulphuric acid is 89.3 g and the weight of sodium hydroxide is 96 g.

Therefore, convert 89.3 g of [tex]H_2SO_4[/tex] and 96.0 g of [tex]NaOH[/tex] into moles by using the formula,

[tex]\rm{Moles=\dfrac{Given\;weight}{Molar Mass}}[/tex]

Known Quantity:

[tex]\rm{Molar\; mass\; of\;} H_2SO_4 = 98.1\; g/mol[/tex]

[tex]\rm{Molar\; mass\; of\; NaOH = 40.0\; g/mol}[/tex]

Hence,

[tex]\begin{aligned}\rm{Moles\;of\;H_2SO_4}&=\dfrac{89.3}{98.1}\\&=0.910\end{aligned}[/tex]

 

[tex]\begin{aligned}\rm{Moles\;of\;NaOH}&=\dfrac{96}{40}\\&=2.40\end{aligned}[/tex]

[tex]\rm{Theoretical \;molar\; ratio} = \dfrac{2\; moles\; NaOH}{1\; mole\; H_2SO_4}[/tex]

Therefore,

If 0.91 moles react of [tex]H_2SO_4[/tex] then the number of moles required of [tex]NaOH[/tex] for the reaction will be twice as 0.91 moles.  

Moles required of [tex]NaOH[/tex] is 1.82.

Thus,

The remaining moles of NaOH will be (2.40 - 1.82) that is 0.58.

Now,

The weight of 0.580 moles NaOH will be calculated by the below expression:

[tex]0.580 \;\rm{moles} \times 40.0\;\rm{ g/mol} = 23.2\; g\; of\; NaOH \;will\; be\; left.[/tex]

For more information, please refer to the link:

https://brainly.com/question/20486415?referrer=searchResults

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