You need to calculate the molar mass for Al(CN)3 using the atomic weights for Al, C, and N given on the periodic table.
1 Al (26.98) + 3 C (3 x 12.01) + 3 N (3 x 14.01) = 105.04 g Al(CN)3 / mole
183 g Al(CN)3 x (1 mole Al(CN)3 / 105.04 g Al(CN)3) = 1.74 moles Al(CN)3
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Answer:
0.497 moles/L
Explanation:
The reaction that takes place is:
- C₃H₆O₃ + NaOH → C₃H₅ONa + H₂O
First we c<u>alculate the moles of lactic acid in 0,821 g</u>, using its molar mass:
- 0,821 g ÷ 90 g/mol = 9,122x10⁻³mol lactic acid
<em>One mole of lactic acid reacts with one mole of sodium hydroxide</em> (it is a monoprotic acid), so in 18,34 mL of NaOH solution there are 9,122x10⁻³moles of NaOH:
- 18.34 mL ÷ 1000 = 0,01834 L
- 9,122x10⁻³mol ÷ 0,01834 L = 0.497 moles/L
