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

1 Answer
Stefan V. · Dwayne M.
Dec 5, 2014

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

Explanation:

Starting from the balanced equation

2"Na"_ ((s)) + "H"_ 2"SO"_ (4(aq)) -> "Na"_ 2"SO"_ (4(aq)) + "H"_ (2(g)) uarr2Na(s)+H2SO4(aq)→Na2SO4(aq)+H2(g)↑⏐

one can see that 11 mole of "H"_2"SO"_4H2SO4 needs 22 moles of "Na"Na. The number of moles of H_2SO_4H2SO4 can be determined from

n_ ("H"_ 2"SO"_ 4) = c_("H"_2"SO"_4) * V_ ("H"_2"SO"_4)nH2SO4=cH2SO4⋅VH2SO4

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.

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