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Question

A scientist is investigating the following reaction: 2K + MgBr2 --> 2KBr + Mg If they started with 34 g of K and had an unlimited supply of MgBr2 and produced 6 g of Mg, what is their percent yield of Mg? (Round your answer to 1 decimal place).​

Answer 1

Hello, Kaycie,

2K + MgBr₂ --> 2KBr + Mg

This balanced reaction states we should expect 1 mole of Mg for every 2 moles of potassium, K. The conditions we are given tell us we should assume all of the K reacts, since there is a surplus of magnesium bromide. If we calculate moles K, we can take 1/2 that value and use that as our predicted moles of Mg.

Moles K: 34g/39.1 g./mole = 0.870 moles of K

Since the balanced equation says it takes 2 moles of potassium to make 1 mole Mg, we can say (1/2)*(0.870 moles K) = 0.435 moles Mg.

Convert moles Mg to grams by multiplying by the molar mass of magnesium: (0.435 moles)*(24.3 g/mole) = 10.6 grams Mg. We actually got 6 grams, sothe percent theorectical yield is 6/10.6 = 56.8%. 57% with 2 sig figs.

Bob