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Calculate the '''yield''' of [[Magnesium Oxide]] given 48g of [[Magnesium]] and excess [[Oxygen]].
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Find the '''yield''' of [[Carbon DioxideWater]] when 32g 320g of [[Methane]] [[Chemical reaction|reacts]] with excess [[Oxygen]].
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Find the '''yield''' of [[Sodium Chloride]] when 80g 7.3 of [[Sodium HydroxideHydrochloric Acid]] [[Chemical Reaction|reactreacts]] with 73g of excess [[Hydrochloric AcidSodium Hydroxide]].
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Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].
M<sub>r</sub> of 2Mg Mg = 48g24g
M<sub>r</sub> of 2MgO MgO = 80g40g
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M<sub>r</sub> of CH<sub>4</sub> = 16g
M<sub>r</sub> of COH<sub>2</sub> O = 44g18g
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State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
1 [[mole]] of CH<sub>4</sub> is needed for every 1 2 [[mole]] of COs H<sub>2</sub>O
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No. [[Mole]]s = <math>\frac{48}{24}</math>
No. [[Mole]]s = 2 Molemol
Therefore 1 2 [[mole]] s of O<sub>2</sub> MgO is neededproduced.
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No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
No. [[Mole]]s = <math>\frac{32320}{16}</math> No. [[Mole]]s = 20 mol
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No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
No. [[Mole]]s = <math>\frac{207.3}{4036.5}</math>
No. [[Mole]]s = 0.5 Mole2 mol
Therefore 0.5 2 [[mole]] of HCl [[Sodium Chloride]] is neededproduced.
0.5 2 [[mole]] of HCl NaCl = 1811.25g7g
'''Yield''' = 11.7g
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