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Step by Step Instructions for Solving a Limited Excess Reagent Stoichiometry Problem. By: Joseph Wheeler

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How many grams of Sodium Sulfide is produced when 9.36 grams of Sodium is reacted with 17.3 grams of Sulfur? Which is the limiting reagent? Which is the excess reagent? How much excess will there be? How many grams of Sodium Sulfide is produced when 9.36 grams of Sodium is reacted with 17.3 grams of Sulfur? Which is the limiting reagent? Which is the excess reagent? How much excess will there be?

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Step 1. Write a complete and balanced chemical equation. 2Na + S Na 2 S 2Na + S Na 2 S

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Step 2. Draw a column after each chemical. 2Na + S Na 2 S 2Na + S Na 2 S

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Step 3. Write the given amounts into the correct columns. 2Na + S Na 2 S 9.36g 17.3g 2Na + S Na 2 S 9.36g 17.3g

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Step 4. Convert the given amounts into moles. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) m = (17.3g/1) x (m/32.066g) m = Na + S Na 2 S (9.36g/1) x (m/22.99g) m = (17.3g/1) x (m/32.066g) m = 0.504

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Step 5a. In each of the other columns write moles of given x a fraction. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (0.407m/1) x m = (0.540m/1) x (17.3g/1) x (m/32.066g) (0.540m/1) x m = Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (0.407m/1) x m = (0.540m/1) x (17.3g/1) x (m/32.066g) (0.540m/1) x m = 0.540

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Step 5b. The numerator of the fraction is the coefficient of that column. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/ (0.407m/1) x (1/ m = (0.540m/1) x (2/ (17.3g/1) x (m/32.066g) (0.540m/1) x (1/ m = Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/ (0.407m/1) x (1/ m = (0.540m/1) x (2/ (17.3g/1) x (m/32.066g) (0.540m/1) x (1/ m = 0.540

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Step 5c. The denominator of the fraction is the coefficient of the given column. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 0.540

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Step 5d. Do the math and label as moles. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = m = m = (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 1.08 m = m = Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = m = m = (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 1.08 m = m = 0.540

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Step 6. Convert all moles to grams. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = m = m = (0.204m/1) x (32.066g/m) (0.204m/1) x (78.046g/m) g = 6.54 g = 15.9 (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 1.08 m = m = (1.08m/1) x (22.99g/m) (0.540m/1) x g/m) g = 24.8 g = Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = m = m = (0.204m/1) x (32.066g/m) (0.204m/1) x (78.046g/m) g = 6.54 g = 15.9 (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 1.08 m = m = (1.08m/1) x (22.99g/m) (0.540m/1) x g/m) g = 24.8 g = 42.1

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Step 7. Verify the law of conservation of mass. 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = m = m = (0.204m/1) x (32.066g/m) (0.204m/1) x (78.046g/m) g = 6.54 g = g g = 15.9 (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 1.08 m = m = (1.08m/1) x (22.99g/m) (0.540m/1) x g/m) g = 24.8 g = g g = 42.1g 2Na + S Na 2 S (9.36g/1) x (m/22.99g) (0.407m/1) x (1/2) (0.407m/1) x (1/2) m = m = m = (0.204m/1) x (32.066g/m) (0.204m/1) x (78.046g/m) g = 6.54 g = g g = 15.9 (0.540m/1) x (2/1) (17.3g/1) x (m/32.066g) (0.540m/1) x (1/1) m = 1.08 m = m = (1.08m/1) x (22.99g/m) (0.540m/1) x g/m) g = 24.8 g = g g = 42.1g

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15.9 grams of Sodium Sulfide is produced when 9.36 grams of Sodium reacts with 17.3 grams of Sulfur. The limiting reagent is Na. The excess reagent is S. There is 10.8 grams of excess Sulfur = grams of Sodium Sulfide is produced when 9.36 grams of Sodium reacts with 17.3 grams of Sulfur. The limiting reagent is Na. The excess reagent is S. There is 10.8 grams of excess Sulfur = 10.8

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