My theoretical yield of beryllium chloride was 10.7 grams in #Be + 2HCl -> BeCl_2 + H_2#. If my actual yield was 4.5 grams, what was my percent yield?

1 Answer
Mar 19, 2016

Answer:

#"% yield" = 42%#

Explanation:

You're dealing with a single replacement reaction in which beryllium metal, #"Be"#, reacts with hydrochloric acid, #"HCl"#, to form aqueous beryllium chloride, #"BeCl"_2#, and hydrogen gas, #"H"_2#, which bubbles out of solutions.

#"Be"_text((s]) + 2"HCl"_text((aq]) -> "BeCl"_text(2(aq]) + "H"_text(2(g]) uarr#

The theoretical yield of beryllium chloride is calculated by taking into account the stoichiometric coefficients of the reactants and of beryllium chloride.

More specifically, you know that one mole of beryllium metal reacts with two moles of hydrochloric acid to form one mole of aqueous beryllium chloride.

This corresponds to a reaction that has a #100%# yield. In your case, the theoretical yield is said to be equal to #"10.7 g"#. This corresponds to the #1:1# mole ratio that exists between beryllium metal and beryllium chloride.

However, you collect #"4.5 g"# of beryllium chloride, your actual yield.

The percent yield of a chemical reaction is defined as

#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you should theoretically get" xx 100color(white)(a/a)|)))#

Plug in your values to get

#"% yield" = (4.5 color(red)(cancel(color(black)("g"))))/(10.7color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)42%color(white)(a/a)|)))#

The answer is rounded to two sig figs, the number of sig figs you have for the actual yield of the reaction.