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#3 part C

1 15126791 HMAC-d3bl02tefddo615328d1bf182619cfbde10301 Hybrid Chem 1411 Sp 19 CAwe rement 15 Chap 2 Linting Re.ctares d Pwcen
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Answer #1

Molar mass of sodium bicarbonate = 84 g/mol

Moles of NaHCO3 in 1.5 g = 1.5/84 = 0.0178 mol

Molar mass of citric acid = 192.124 g/mol

Moles of citric acid in 1.5 g = 1.5/192.124 = 0.0078 mol

2 moles of sodium bicarbonate react with 1 mole of citric acid.

0.0178 mol of sodium bicarbonate reacts with 0.0178/2 = 0.0089 mol of citric acid

Available citric acid is only 0.0078 mol

0.0078 mol of citric acid reacts with 0.0078*2 = 0.0156 moles of sodium bicarbonate.

Excess sodium bicarbonate = 0.0178-0.0156 = 0.0022 mol

Mass of 0.0022 mol of NaHCO3 = 0.0022*84 = 0.1848 g

Answer : 0.1848 g

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