How many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl Solution

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Solution : 1) write a balanced equation 2) convert everything you were given to moles. 3) determine limiting reagent. 4) convert from moles of one species to moles of another 5) convert moles to grams.... **** 1 **** PbO + NaCl + H2O -----> Pb(OH)Cl + NaOH.... is balanced **** 2 **** we do this step because the coefficients of the balanced equation are in mole ratios. not grams or liters or whatever. and we will use those coefficients to convert back and forth between species. in the next step... moles = mass /mw right? or mass x 1/mw....same thing... mw Pb(OH)Cl = 1x207.2 + 1x16.0+ 1x1.01 + 1x35.45 = 259.7 g/mole moles Pb(OH)Cl = 10.0 g x ( 1 mole / 259.7 g) = 0.0385 moles **** 3 **** determine limiting reagent...although we'll skip this step, I left it in because you need to do this for many other problems. The idea is that sometimes you don't have enough of one reactant to consume all of the other. So that reactant is limiting. for example, if you had been given 0.5 moles of PbO, 1 mole NaCl and 1 mole H2O to react, then since you need 1 mole of PbO to react consume all the others and you have less than that, PbO is limiting. In any case, it isn't necessary for this particular problem. **** 4 **** from balanced equation.... 1 mole Pb(OH)Cl = 1 mole PbO so... 0.0385 moles Pb(OH)Cl x (1 mole PbO / 1 mole Pb(OH)Cl ) = 0.0385 PbO similarly from the coefficients of the balanced equation 1 mole Pb(OH)Cl = 1 mole NaOH 1 mole Pb(OH)Cl = 1 mole H2O converting .... 0.0385 moles Pb(OH)Cl x (1 mole NaOH / 1 mole Pb(OH)Cl ) = 0.0385 NaOH 0.0385 moles Pb(OH)Cl x (1 mole H2O / 1 mole Pb(OH)Cl ) = 0.0385 H2O **** 5 **** convert moles back to grams moles = mass / mw mass = moles x mw right? mw PbO = 207.2+16 = 223.3 g/mole mw NaOH = 23.0+16.0+1.0 = 40.0 g/mole mw H2O = 2 x 1.0 + 1 x 16 = 18.0 g/mole mass PbO = 0.0385 moles PbO x (223.3 g PbO / mole PbO) = 8.59 g mass NaOH = 0.0385 moles NaOH x (40.0 g NaOH / mole NaOH) = 1.54 g mass H2O = 0.0385 moles H2O x (18.0 g H2O / mole H2O) = 0.693 g Please rate my answer.

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