Question

Calculate the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess. Also write and balance an equation.

Answer 1

The balanced chemical equation for the production of chromium metal from the reaction of chromium(ll) nitrate reacts with a strip of zinc is:

3 Zn + 2 Cr(NO₃)₃ → 2 Cr + 3 Zn(NO₃)₂

This is a redox reaction, which is a chemical reaction in which one or more electrons are transferred between the reagents, causing a change in their oxidation states. In the proposed reaction, Cr oxidation state goes from +3 to 0, becoming metallic chromium, while Zn goes from being Zn⁰ to Zn²⁺.

The mass of chromium metal produced in the above reaction will be,

425.0 mL x $\frac{1 L}{1000 mL}$ x  $\frac{0.25 mol Cr(NO_{3})_{3} }{1 L}$ x $\frac{2 mol Cr }{2 mol Cr(NO_{3})_{3} }$ x $\frac{51.9961 g Cr}{1 mol Cr}$ = 5.52 g

So, the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess is 5.52 g of Cr.