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

Dec 5, 2014

$\text{210 g}$ of $\text{Na}$.

#### Explanation:

Starting from the balanced equation

$2 {\text{Na"_ ((s)) + "H"_ 2"SO"_ (4(aq)) -> "Na"_ 2"SO"_ (4(aq)) + "H}}_{2 \left(g\right)} \uparrow$

one can see that $1$ mole of ${\text{H"_2"SO}}_{4}$ needs $2$ moles of $\text{Na}$. The number of moles of ${H}_{2} S {O}_{4}$ can be determined from

${n}_{{\text{H"_ 2"SO"_ 4) = c_("H"_2"SO"_4) * V_ ("H"_2"SO}}_{4}}$

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 $N a$ is $22.99 \text{g/mol}$, the mass of $\text{Na}$ 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.