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Solution stoich.

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Presentation on theme: "Solution stoich."— Presentation transcript:

1 Solution stoich

2 c = n v concentration C. Solution Stoichiometry
use to perform calculations concentration c = n v

3 w g 2 KOH(aq) + 1 H2SO4(aq)  1 K2SO4(aq) + 2 H2O(l) v=? c= 14.8 mol/L
Example 1 What volume of 14.8 mol/L KOH is needed to react completely with 1.50 L of 12.9 mol/L sulphuric acid? w g 2 KOH(aq) + 1 H2SO4(aq) 1 K2SO4(aq) + 2 H2O(l) v=? c= 14.8 mol/L v = 1.50 L c = 12.9 mol/L n = x 2 1 = mol n = cv = 12.9 mol/L x 1.50 L = mol v = n c = mol 14.8 mol/L = L = 2.61 L


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