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(d) If 48.0 g of NaCl react with 19.0 g of H₂SO4, what mass of Na2SO4 will be produced?​

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Answer:

the mass of Na2SO4 produced is 27.6 g.

Step-by-step explanation:

determine the balanced chemical equation for the reaction between NaCl and H₂SO4, and then use stoichiometry to calculate the mass of Na2SO4 produced.

The balanced chemical equation for the reaction is:

2 NaCl + H₂SO4 → Na2SO4 + 2 HCl

This equation tells us that 2 moles of NaCl react with 1 mole of H₂SO4 to produce 1 mole of Na2SO4.

First, we need to determine the number of moles of NaCl and H₂SO4 we have:

moles of NaCl = mass / molar mass = 48.0 g / 58.44 g/mol = 0.821 mol

moles of H₂SO4 = mass / molar mass = 19.0 g / 98.08 g/mol = 0.194 mol

Since H₂SO4 is the limiting reactant (it is present in less amount than required for complete reaction), we need to use the number of moles of H₂SO4 to calculate the number of moles of Na2SO4 produced:

moles of Na2SO4 = moles of H₂SO4 / 1 x (1 mole Na2SO4 / 1 mole H₂SO4) = 0.194 mol x (1/1) = 0.194 mol

Finally, we can use the molar mass of Na2SO4 to calculate the mass of Na2SO4 produced:

mass of Na2SO4 = moles of Na2SO4 x molar mass = 0.194 mol x 142.04 g/mol = 27.6 g

Therefore, the mass of Na2SO4 produced is 27.6 g.

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