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Given the following reaction: H2SO42LiOH=Li2SO42H20, what ma of water i produced from 10 g of ulfuric acid?

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Answer:

The balanced equation for the reaction is:

H2SO4 + 2 LiOH → Li2SO4 + 2 H2O

From the balanced equation, it can be determined that 2 moles of water are produced per mole of sulfuric acid. To find the mass of water produced from 10 g of sulfuric acid, we need to convert 10 g to moles and then multiply by the number of moles of water produced per mole of sulfuric acid.

The molar mass of sulfuric acid is 98 g/mol, so 10 g of sulfuric acid is 10 g / 98 g/mol = 0.102 mol.

Thus, 0.102 mol * 2 moles of water / 1 mole of sulfuric acid = 0.204 moles of water.

And finally, the mass of water produced is 0.204 moles * 18 g/mol (the molar mass of water) = 3.672 g.

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