If 0.953 g of copper was actually recovered at the end of the entire experiment, what would be the percent yield?

Here is the full question:

Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

1.000 g copper wire is reacted with 10 mL of concentrated nitric acid (16M). If 0.953 g of copper was actually recovered at the end of the entire experiment, what would be the percent yield?

95.30 %

Explanation:

equation for the reaction is given as:

Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

1.00 g of Cu =

= 0.01574 mol of Cu

10 mL of 16 M HNO3 =

= 0.16 mol of HNO3

However, from both variables; we can arrive at a conclusion that Cu is the limiting reagent.

therefore % yield = = 4.70 %

Percentage yield =

Percentage yield = %

Percentage yield = 0.953 × 100%

Percentage yield = 95.30 %

the answer is minerals.

answer is: naturally occurring solid, inorganic compound or element is mineral.

minerals usually have crystalline form, that means than crystal constituents (atoms, molecules or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

for example magnetite is a mineral of iron oxide (fe₂o₃).

calcite is a carbonate mineral and the most stable polymorph of calcium carbonate (caco₃).

dolomite is an   carbonate mineral composed of calcium magnesium carbonate (camg(co₃)₂).