How many grams of BaCl₂ are formed when 35.00 mL of 0.00237 M Ba(OH)₂ reacts with excess Cl₂ gas? 2 Ba(OH)₂(aq)+ 2 Cl₂(g)→ Ba(OCl)₂(aq)+ BaCl₂(s)+ 2 H₂O(l) A)0.00864 g B)0.0173 g C)0.0346 g D)0.0829 g

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Quizplus · Accepted Answer

First, calculate moles of Ba(OH)₂: 0.00237 M × 0.035 L = 8.295 × 10⁻⁵ moles. From the balanced equation, 1 mole of Ba(OH)₂ produces 1 mole of BaCl₂. Therefore, 8.295 × 10⁻⁵ moles of BaCl₂ are formed. The molar mass of BaCl₂ is 137.33 + 2(35.45) = 208.23 g/mol. Thus, mass = 8.295 × 10⁻⁵ moles × 208.23 g/mol = 0.0173 g.

How Many Grams of BaCl2 Are Formed When 35

How Many Grams of BaCl₂ Are Formed When 35