what is the volume of 15.8 M nitric acid is needed to ensure 1.00 g sample copper completely reacts
M(Cu) = 63.54 g/mol
n(Cu) =1.0063.54=0.0157 mol= \frac{1.00}{63.54} = 0.0157 \;mol=63.541.00=0.0157mol
According to the reaction equation:
n(HNO3) = 4n(Cu) =4×0.0157=0.0629 mol= 4 \times 0.0157 = 0.0629 \;mol=4×0.0157=0.0629mol
Proportion:
15.8 mol – 1000 mL
0.0629 mol – x
x=0.0629×100015.8=3.98 mL≈4 mLx = \frac{0.0629 \times 1000}{15.8} = 3.98 \; mL ≈ 4 \;mLx=15.80.0629×1000=3.98mL≈4mL
Answer: 4 mL
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