QuestionDecember 15, 2025

6. How many mols of CO_(2) are formed with 4.87 grams of baking soda? Click on the letter in the formula to insert a number. [A]x([A])/([A])times ([C])/([D])times ([D])/([C])=

6. How many mols of CO_(2) are formed with 4.87 grams of baking soda? Click on the letter in the formula to insert a number. [A]x([A])/([A])times ([C])/([D])times ([D])/([C])=
6. How many mols of CO_(2) are formed with 4.87 grams of baking soda? Click on the letter in the formula to insert a number. [A]x([A])/([A])times ([C])/([D])times ([D])/([C])=

Solution
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Answer

0.0580\ \mathrm{mol\ CO_{2}} Explanation 1. Identify the molar mass of baking soda (NaHCO₃) Molar mass: Na (22.99) + H (1.01) + C (12.01) + 3O (3 \times 16.00) = 84.01\ \mathrm{g/mol}. 2. Convert grams to moles of NaHCO₃ **Formula:** n = \frac{m}{M} n = \frac{4.87}{84.01} \approx 0.05796\ \mathrm{mol}. 3. Determine mole ratio between NaHCO₃ and CO₂ Reaction: NaHCO_{3} \rightarrow Na_{2}CO_{3} + CO_{2} + H_{2}O produces 1 mol CO₂ per 1 mol NaHCO₃. Ratio = 1:1. 4. Calculate moles of CO₂ Moles of CO₂ = 0.05796 \times 1 = 0.05796\ \mathrm{mol}.

Explanation

1. Identify the molar mass of baking soda (NaHCO₃) <br /> Molar mass: $Na (22.99) + H (1.01) + C (12.01) + 3O (3 \times 16.00) = 84.01\ \mathrm{g/mol}$.<br /><br />2. Convert grams to moles of NaHCO₃ <br /> **Formula:** $n = \frac{m}{M}$ <br /> $n = \frac{4.87}{84.01} \approx 0.05796\ \mathrm{mol}$.<br /><br />3. Determine mole ratio between NaHCO₃ and CO₂ <br /> Reaction: $NaHCO_{3} \rightarrow Na_{2}CO_{3} + CO_{2} + H_{2}O$ produces 1 mol CO₂ per 1 mol NaHCO₃. Ratio = 1:1.<br /><br />4. Calculate moles of CO₂ <br /> Moles of CO₂ = $0.05796 \times 1 = 0.05796\ \mathrm{mol}$.
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