Contents
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
|
Calculate the yield of Magnesium Oxide given 48g of Magnesium and excess Oxygen. |
Find the yield of Carbon Dioxide when 32g of Methane reacts with excess Oxygen. |
Find the yield of Sodium Chloride when 80g of Sodium Hydroxide react with 73g of Hydrochloric Acid. |
|
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 |
|
Find the Relative Formula Mass of the relevant reactants and product. Mr of 2Mg = 48g Mr of 2MgO = 80g |
Find the Relative Formula Mass of the relevant reactants and product. Mr of CH4 = 16g Mr of CO2 = 44g |
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 |
|
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{48}{24}\) No. Moles = 2 Mole Therefore 1 mole of O2 is needed. 1 mole of O2 = 32g |
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{32}{16}\) No. Moles = 2 Mole Therefore 4 moles of O2 are needed. 4 moles of O2 = 128g |
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{20}{40}\) No. Moles = 0.5 Mole Therefore 0.5 mole of HCl is needed. 0.5 mole of HCl = 18.25g |