Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O. Suppose 17. g of sulfuric acid is mixed with 7.07 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

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Explanation:

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Maas of sodium sulphate produced = moles×molar mass=0.06×142=8.52 gmoles\times molar\ mass=0.06\times 142=8.52\ gmoles×molar mass=0.06×142=8.52 g

Explanation:

Reaction between them is given by-

2NaOH  +H2SO4 → Na2SO4  +2H2O2NaOH\ \ +H_2SO_4\ \to\ Na_2SO_4\ \ +2H_2O2NaOH  +H2​SO4​ → Na2​SO4​  +2H2​O

Molar mass of H2SO4=98 g mole−1H_2SO_4=98\ g\ mole^{-1}H2​SO4​=98 g mole−1

Moles of sulphuric acid taken = 5.8898=0.06 mole\frac{5.88}{98}=0.06\ mole985.88​=0.06 mole

Moles of sodium hydroxide taken = 6.240=0.155 mole\frac{6.2}{40}=0.155\ mole406.2​=0.155 mole

here sulphuric acid is limiting agent so

reaction will proceed according to sulphuric acid

0.06 mole HSO4 will produce same amount of  Na2SO40.06\ mole\ H_SO_4\ will \ produce\ same\ amount\ of \ \ Na_2SO_4\\0.06 mole HS​O4​ will produce same amount of  Na2​SO4​

the answer for sodium is 0.186

the answer for rubidium is 0.251