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 would be c

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.