Calculate the theoretical yield of lead bromide if 0.357 g of potassium bromide and excess lead nitrate are reacted according to the following reaction:2 KBr(aq) + Pb(NO3)2(aq) --> PbBr2(s) + 2 KNO3(aq)Molar MassesKBr = 119.00 g mol-1Pb(NO3)2 = 331.22 g mol-1PbBr2 = 367.00 g mol-1KNO3 = 101.11 g mol-1Question 1Answer0.1790.551 g0.79 g0.396 g0.198 g
Question
Calculate the theoretical yield of lead bromide if 0.357 g of potassium bromide and excess lead nitrate are reacted according to the following reaction:2 KBr(aq) + Pb(NO3)2(aq) --> PbBr2(s) + 2 KNO3(aq)Molar MassesKBr = 119.00 g mol-1Pb(NO3)2 = 331.22 g mol-1PbBr2 = 367.00 g mol-1KNO3 = 101.11 g mol-1Question 1Answer0.1790.551 g0.79 g0.396 g0.198 g
Solution
To calculate the theoretical yield of PbBr2, we first need to determine the number of moles of KBr that are present.
Step 1: Calculate the number of moles of KBr Number of moles = mass / molar mass Number of moles of KBr = 0.357 g / 119.00 g mol-1 = 0.003 mol
Step 2: From the balanced chemical equation, we can see that the ratio of KBr to PbBr2 is 2:1. Therefore, the number of moles of PbBr2 that can be formed from 0.003 mol of KBr is 0.003 mol / 2 = 0.0015 mol.
Step 3: Calculate the mass of PbBr2 that can be formed Mass = number of moles * molar mass Mass of PbBr2 = 0.0015 mol * 367.00 g mol-1 = 0.551 g
So, the theoretical yield of PbBr2 is 0.551 g. Therefore, the correct answer is 0.551 g.
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