Question

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

Answer 1

`"210 g"` of `"Na"`.

Explanation:

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.