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 %

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

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