Question

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 into their beakers. The reaction between the copper in the mixture and the HNO3 is represented by the equation above. The students observed that a brown gas was released from the beakers and that the solutions turned blue, indicating the formation of Cu 2. The solutions were then diluted with distilled water to known volumes. The students determined that the reaction produced 0.010 mol of Cu(NO3)2. Based on the measurement, what was the percent of Cu by mass in the original 2.00 g sample of the mixture

Answer 1

Answer:

Percentage mass of copper in the sample = 32%

Explanation:

Equation of the reaction producing Cu(NO₃) is given below:

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

From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.

Molar mass of copper = 64 g/mol

mass of copper = number of moles * molar mass

mass of copper = 0.01 mol * 64 g/mol = 0.64 g

Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%

Percentage mass of copper in the sample = 32%