Difference between revisions of "Yield (Chemistry)"
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Calculate the '''yield''' of [[Magnesium Oxide]] given 48g of [[Magnesium]] and excess [[Oxygen]]. | Calculate the '''yield''' of [[Magnesium Oxide]] given 48g of [[Magnesium]] and excess [[Oxygen]]. | ||
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| − | Find the '''yield''' of [[ | + | Find the '''yield''' of [[Water]] when 320g of [[Methane]] [[Chemical reaction|reacts]] with excess [[Oxygen]]. |
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| − | Find the '''yield''' of [[Sodium Chloride]] when | + | Find the '''yield''' of [[Sodium Chloride]] when 7.3 of [[Hydrochloric Acid]] [[Chemical Reaction|reacts]] with excess [[Sodium Hydroxide]]. |
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Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]]. | Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]]. | ||
| − | M<sub>r</sub> of | + | M<sub>r</sub> of Mg = 24g |
| − | M<sub>r</sub> of | + | M<sub>r</sub> of MgO = 40g |
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M<sub>r</sub> of CH<sub>4</sub> = 16g | M<sub>r</sub> of CH<sub>4</sub> = 16g | ||
| − | M<sub>r</sub> of | + | M<sub>r</sub> of H<sub>2</sub>O = 18g |
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State the [[ratio]] of [[mole]]s of each [[chemical]] needed. | State the [[ratio]] of [[mole]]s of each [[chemical]] needed. | ||
| − | 1 [[mole]] of CH<sub>4</sub> is needed for every | + | 1 [[mole]] of CH<sub>4</sub> is needed for every 2 [[mole]]s H<sub>2</sub>O |
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No. [[Mole]]s = <math>\frac{48}{24}</math> | No. [[Mole]]s = <math>\frac{48}{24}</math> | ||
| − | No. [[Mole]]s = 2 | + | No. [[Mole]]s = 2 mol |
| − | Therefore | + | Therefore 2 [[mole]]s of MgO is produced. |
| − | + | 2 [[mole]]s of MgO = 40x2 | |
| + | |||
| + | '''Yield''' = 80g | ||
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No. [[Mole]]s = <math>\frac{Mass}{M_r}</math> | No. [[Mole]]s = <math>\frac{Mass}{M_r}</math> | ||
| − | No. [[Mole]]s = <math>\frac{ | + | No. [[Mole]]s = <math>\frac{320}{16}</math> |
| + | |||
| + | No. [[Mole]]s = 20 mol | ||
| − | + | Therefore 40 [[mole]]s of H<sub>2</sub>O are produced. | |
| − | + | 40 [[mole]]s of CO<sub>2</sub> = 18x40 | |
| − | + | '''Yield''' = 720g | |
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No. [[Mole]]s = <math>\frac{Mass}{M_r}</math> | No. [[Mole]]s = <math>\frac{Mass}{M_r}</math> | ||
| − | No. [[Mole]]s = <math>\frac{ | + | No. [[Mole]]s = <math>\frac{7.3}{36.5}</math> |
| − | No. [[Mole]]s = 0. | + | No. [[Mole]]s = 0.2 mol |
| − | Therefore 0. | + | Therefore 0.2 [[mole]] of [[Sodium Chloride]] is produced. |
| − | 0. | + | 0.2 [[mole]] of NaCl = 11.7g |
| + | '''Yield''' = 11.7g | ||
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Revision as of 17:49, 12 January 2019
Key Stage 4
Meaning
Yield is the mass of product in a chemical reaction.
About Yield
- The theoretical yield of product form a chemical reaction can be calculated using the balanced symbol equation and the relative formula masses of all the molecules involved.
- The experimental yield from a chemical reaction is often not as large as the theoretical yield because not all of the reactants will react. This experimental yield is often expressed as a percentage of the theoretical yield and is known as the Percentage Yield.
Examples
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Calculate the yield of Magnesium Oxide given 48g of Magnesium and excess Oxygen. |
Find the yield of Water when 320g of Methane reacts with excess Oxygen. |
Find the yield of Sodium Chloride when 7.3 of Hydrochloric Acid reacts with excess Sodium Hydroxide. |
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Write the Balanced Symbol Equation 2Mg + O2 → 2MgO |
Write the Balanced Symbol Equation CH4 + 2O2 → 2H2O + CO2 |
Write the Balanced Symbol Equation NaOH + HCl → NaCl + H2O |
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Find the Relative Formula Mass of the relevant reactants and product. Mr of Mg = 24g Mr of MgO = 40g |
Find the Relative Formula Mass of the relevant reactants and product. Mr of CH4 = 16g Mr of H2O = 18g |
Find the Relative Formula Mass of the relevant reactants and product. Mr of NaOH = 40g Mr of HCl = 36.5g Mr of NaCl = 58.5g |
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Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{48}{24}\) No. Moles = 2 mol Therefore 2 moles of MgO is produced. 2 moles of MgO = 40x2 Yield = 80g |
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{320}{16}\) No. Moles = 20 mol Therefore 40 moles of H2O are produced. 40 moles of CO2 = 18x40 Yield = 720g |
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{7.3}{36.5}\) No. Moles = 0.2 mol Therefore 0.2 mole of Sodium Chloride is produced. 0.2 mole of NaCl = 11.7g Yield = 11.7g |