How many grams of sodium can react with 750 mL of a 6.0 M solution of sulfuric acid, H2SO4? Na + H2SO4 Na2SO4 + H2

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Answer 1 · 2014-12-05T23:10:40

$"210 g"$ of $"Na"$ .

Starting from the balanced equation

$2"Na"_ ((s)) + "H"_ 2"SO"_ (4(aq)) -> "Na"_ 2"SO"_ (4(aq)) + "H"_ (2(g)) uarr$

one can see that $1$ mole of $"H"_2"SO"_4$ needs $2$ moles of $"Na"$ . The number of moles of $H_2SO_4$ can be determined from

$n_ ("H"_ 2"SO"_ 4) = c_("H"_2"SO"_4) * V_ ("H"_2"SO"_4)$

$n_ ("H"_ 2"SO"_ 4) = "6 mol/"cancel("L") * 0.75 cancel("L") = "4.5 moles"$

This means that we need a minimum of

$n_("Na") = 2 * n_("H"_2"SO"_4) = 2 * 4.5 = "9 moles"$

Knowing that the molar mass of $Na$ is $22.99 "g/mol"$ , the mass of $"Na"$ will be

$m_("Na") = n_("Na") * "22.99 g/mol"$

$= 9 cancel("moles") * "22.99 g/"cancel("mol") = "210 g"$

The answer is rounded to two significant digits.

How many grams of sodium can react with 750 mL of a 6.0 M solution of sulfuric acid, H2SO4? __ Na + __ H2SO4 __ Na2SO4 + __ H2