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vaieri [72.5K]
3 years ago
10

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

ater . Suppose 91. g of sulfuric acid is mixed with 116. g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to significant digits.
Chemistry
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

maximum mass of sodium sulphate=132g

Explanation:

firstly balance the chemical reaction,

H_2 SO_4_aq+2NaOH_s=Na_2 SO_4_aq+2H_2 O_l

molecular weight of H2SO4=98g/mole;

Molecular weight of NaOH=40g/mole;

number of mole of H2SO4 given=91g/(98g/mole)=0.93mole;

number of mole of NaOH given=116g/(40g/mole)=2.9mole

moles are:

number of mole of H2SO4=0.93

number of mole of NaOH=2.9

From the balanced equation,

1mole of H2SO4 reacts with 2 moles of NaOH;

hence ;

0.93 mole of H2SO4 will react with 2*0.93 moles of NaOH means

moles of NaOH that reacts wih H2SO4=1.86mole but we have given 2.9 mole therefore

NaOH will be excess reagent and H2SO4 will be the limitting reagent

and mass of product depends on limitting reagent i.e mass of H2SO4.

1mole of H2SO4 produces 1 mole of Na2SO4;

0.93 mole produces 0.93 mole of Na2SO4;

maximum mass of Na2SO4 produces=mole*molecular weight;

Molecular weight of Na2SO4=142g/mole;

maximum mass of Na2SO4 produces=0.93*142=132g;

maximum mass ofNa2SO4 produces=132g;

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\large \boxed{\text{-41.2 kJ/mol}}

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\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

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