Calculate the amount of the substance and the mass of water that is formed during the neutralization of NaOH with a mass of 4 g: a) hydrochloric acid b) sulfuric acid c) phosphoric acid Indicate the type of reaction
m (NaOH) = 4 g
n (H2O) -?
m (H2O) -?
1) n (NaOH) = m / M = 4/40 = 0.1 mol;
2) NaOH + HCl => NaCl + H2O;
3) n (H2O) = n (NaOH) = 0.1 mol;
4) m (H2O) = n * M = 0.1 * 18 = 1.8 g;
5) 2NaOH + H2SO4 => Na2SO4 + 2H2O;
6) n (H2O) = n (NaOH) = 0.1 mol;
7) m (H2O) = n * M = 0.1 * 18 = 1.8 g;
8) 3NaOH + H3PO4 => Na3PO4 + 3H2O;
9) n (H2O) = n (NaOH) = 0.1 mol;
10) m (H2O) = n * M = 0.1 * 18 = 1.8 g.
Answer: The amount of substance H2O is 0.1 mol; weight – 1.8 g.
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