A student obtained 3.7g of methyl cinnamate (162g/mol) from 6.5g of cinnamic acid (148g/mol) and 2.0g of methanol (32g/mol). Determine which reagent is limiting and calculate the percentage yield achieved by the student.

Reaction equation:cinnamic acid + methanol --> methyl cinnamate (+ water).Determine the limiting reagent:Calculate the number of moles of each reagent using the triangle g/(GFM)*(n).n = g/GFM cinnamic acid: n = 6.5/148 = 0.044 molesmethanol:n = 2.0/32 = 0.063 molesThere are fewer moles of cinnamic acid which will therefore be used up first, making cinnamic acid the limiting reagent.Calculate % yield:In the reaction, 1 mole of cinnamic acid (limiting reagent) produces 1 mole of methyl cinnamate. Need to calculate expected yield (g).g = n x GFM = 0.044 x 162 = 7.1g% yield = actual yield/expected yield x 100 = 3.7/7.1 x 100 = 52.1%