Chemistry, 06.10.2019 19:30 lilbrown6369

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.

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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

Answer from: LaughterLies

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

Explanation:

Given:

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

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

$H_2SO_4 + 2NaOH \rightarrow Na_2 SO_4 + 2H_2O$

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 H_2SO_4 and 96.0 g of NaOH into moles by using the formula,

$\rm{Moles=\dfrac{Given\;weight}{Molar Mass}}$

Known Quantity:

$\rm{Molar\; mass\; of\;} H_2SO_4 = 98.1\; g/mol$

$\rm{Molar\; mass\; of\; NaOH = 40.0\; g/mol}$

Hence,

$\begin{aligned}\rm{Moles\;of\;H_2SO_4}&=\dfrac{89.3}{98.1}\\&=0.910\end{aligned}$

$\begin{aligned}\rm{Moles\;of\;NaOH}&=\dfrac{96}{40}\\&=2.40\end{aligned}$

$\rm{Theoretical \;molar\; ratio} = \dfrac{2\; moles\; NaOH}{1\; mole\; H_2SO_4}$

Therefore,

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

Moles required of NaOH 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:

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

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Answer from: Quest

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Answer from: Quest

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