Cu(s) + 4 hno3 (aq) --> cu(no3)2 (aq) + 2no2 (g) + 2h2o(l)
each student in a class placed a
2.00 g sample of a mixture of cu and al in a beaker and placed the beaker in a fume hood. the students slowly poured 15.0 ml of 15.8 m hno3 (aq) into their beakers. the reaction between copper in the mixture and hno3 depicted above. the students observed that a brown gas was released from the beakers and that solutions turned blue, indicating the formation of cu2+. the solutions were then diluted with distilled water to known volumes. the students determined that the reaction produced .010 mole fo cu(no3)2. based on the measurement, what was the percent of cu by mass in the original 2.00 gram sample of the mixture? ( provide detailed explanation)
a)16%
b) 32%
c) 64%
d)96%

B) 32%

Explanation:

The data from the problem of which we are interested:

2 gram sample

0.010 moles of Cu(NO₃)₂ produced by the reaction

we calculate the mass of Cu(NO₃)₂ knowing the molecular weight of copper nitrate which is equal to 187.5 grams/mole:

mass = number of moles × molecular weight

mass =  0.010 × 187.5  = 1.875 g

now we devise the following reasoning:

if         187.5 grams of copper nitrate contains 63.5 grams of cooper

then    1.875 grams of copper nitrate contains X grams of cooper

X = (1.875 ×  63.5) /  187.5 = 0.645 grams of cooper

And we have all the values to calculate the percent of Cu by mass in the original mixture:

if in            2 grams of sample we have 0.645 grams of cooper

then in  100  grams of sample we have Y grams of cooper

Y = (100 ×  0.645) / 2 = 31.75 ≈ 32 %

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a. hope i'm right

explanation: