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

Answer:

#"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.