Answer:
17.18 moles of NaCl are in 2,719 mL of a 6.32
Explanation:
1 valence electron in alkali metals.
V(C₄H₆O₃) = 5.00 mL.
d(C₄H₆O₃) = 1.08 g/mL.
m(C₄H₆O₃) = V(C₄H₆O₃) · d(C₄H₆O₃).
m(C₄H₆O₃) = 5.00 mL · 1.08 g/mL.
m(C₄H₆O₃) = 5.4 g.
n(C₄H₆O₃) = m(C₄H₆O₃) ÷ M(C₄H₆O₃).
n(C₄H₆O₃) = 5.4 g ÷ 102 g/mol.
n(C₄H₆O₃) = 0.0529 mol.
n(C₇H₆O₃) = 2.08 g ÷ 138.1 g/mol.
n(C₇H₆O₃) = 0.015 mol; limiting reactant.
From chemical reaction: n(C₄H₆O₃) : n(C₉H₈O₄) = 1 : 1.
n(C₉H₈O₄) = 0.015 mol.
m(C₉H₈O₄) = 0.015 mol · 180.16 g/mol.
m(C₉H₈O₄) = 2.71 g; theoretical yield.
percent yield od aspirine = 2.57 g ÷ 2.71 g · 100% = 94.83%.
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