Problems with stoichiometry?

Help with stoichiometry problems?

• I need help on how to set up these two problems A student used 1.34g of silver to produce silver nitrate. The actual yield is 2.01 . Calculate the percentage yield. To prepare the paint pigment chrome yellow PbCrO4 a student started with 5.52 g of Pb(NO3)2. Actual yield I'd 5.096 , calculate the theoretical yield and percentage yield

• Answer:

  A student used 1.34g of silver to produce silver nitrate. The actual yield is 2.01 . Calculate the percentage yield. First you must write a balanced chemical equation. 2Ag(s) + 2 HNO3(aq) --> 2 AgNO3(aq) + H2(g) Note that you produce 1 mole of AgNO3 for each mole of Ag present. moles Ag = 1.34 g(1 mole Ag)/(107.9 g) = 0.01242 mol Ag Therefore you theoretically produce 0.01242 mol of AgNO3 mass AgNO3 = 0.01242 mol AgNO3(169.87 g/mol) = 2.11 g >> = theoretical yield % yield = [(mass obtained)/(theoretical yield)]*100% = [(2.01 g)/(2.11 g)]*100% = 95.3% To prepare the paint pigment chrome yellow PbCrO4 a student started with 5.52 g of Pb(NO3)2. Actual yield I'd 5.096 , calculate the theoretical yield and percentage yield. Once again you MUST write the chemical equation. Pb(NO3)2(aq) + Na2CrO4(aq) --> PbCrO4(s) + 2 NaNO3(aq) Once again you obtain 1 mole of PbCrO4 from each mole of Pb(NO3)2. Find moles Pb(NO3)2 = (mass Pb(NO3)2)(1 mole)/(molar mass Pb(NO3)2) Convert moles Pb(NO3)2 to moles PbCrO4 (they will be the same) convert moles PbCrO4 to mass. mass PbCrO4 = moles PbCrO4(molar mass PbCrO4) You can either obtain your molar masses from the periodic table or by using Google. You need to convince yourself that you can actually work a problem of this type. Hope this is helpful to you. JIL HIR