Elemental phosphorus occurs as tetratomic molecules, p4. what mass of chlorine gas is needed to react completely with 858 g of phosphorus to form phosphorus pentachloride?
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1) write a balanced equation
2) convert mass, litres, whatever to moles.. this is because coefficients of balanced equations are ratios of moles...
3) determine limiting reagent from coefficients of balanced equation
4) convert from moles of the species you were given to moles of species you need to find using coefficients of balanced equation
5) convert moles back to mass
if you're ever asked to calculate % yield, the result from step (5) is theoretical mass. actual mass is what you measure. % yield = actual / theoretical x 100%
here's the solution...
**** 1 write a balanced equation ****
1 P4 + 10 Cl2 ---> 4 PCl5
**** 2 convert to moles ****
moles = mass / mw
moles P4 = 455 g / (4x30.97g/mole) = 3.672 moles {30.67 is from periodic table}
**** 3 limiting reagent ****
sometimes you're giving two different masses of reactants ---> 2 different values of moles for reactants... if you don't have enough of one reactant to completely consume all of the other according to the mole ratios in the balanced equation, then that reactant is limiting...
For this problem, P4 is limiting because you are given a fixed amount of P4 and allowed to add as much Cl2 as the need to react it all...
**** 4 converts from moles of 1 species to moles of another ****
this is always done based on the limiting reagent. P4 for this problem...Since you know moles P4, and you want to find the mass of Cl2, use the coefficients of the balanced equation to convert moles P4 to moles Cl2.
from step (1), 1 mole P4 reacts with 10 moles Cl2
so ...
3.672 moles P4 x (10 moles Cl2 / 1 mole P4) = 36.72 moles Cl2
**** 5. convert moles back to mass ****
from periodic table, atomic mass Cl = 35.45 g/mole.. so Cl2 = 70.90 g/mole
36.72 moles Cl2 x (70.90 g/mole) = 2603 g
since 455 g has 3 sig figs, and since this was all multiplication and division, your final answer needs to have only 3 sig figs...
mass Cl2 = 2.60 kg
2) convert mass, litres, whatever to moles.. this is because coefficients of balanced equations are ratios of moles...
3) determine limiting reagent from coefficients of balanced equation
4) convert from moles of the species you were given to moles of species you need to find using coefficients of balanced equation
5) convert moles back to mass
if you're ever asked to calculate % yield, the result from step (5) is theoretical mass. actual mass is what you measure. % yield = actual / theoretical x 100%
here's the solution...
**** 1 write a balanced equation ****
1 P4 + 10 Cl2 ---> 4 PCl5
**** 2 convert to moles ****
moles = mass / mw
moles P4 = 455 g / (4x30.97g/mole) = 3.672 moles {30.67 is from periodic table}
**** 3 limiting reagent ****
sometimes you're giving two different masses of reactants ---> 2 different values of moles for reactants... if you don't have enough of one reactant to completely consume all of the other according to the mole ratios in the balanced equation, then that reactant is limiting...
For this problem, P4 is limiting because you are given a fixed amount of P4 and allowed to add as much Cl2 as the need to react it all...
**** 4 converts from moles of 1 species to moles of another ****
this is always done based on the limiting reagent. P4 for this problem...Since you know moles P4, and you want to find the mass of Cl2, use the coefficients of the balanced equation to convert moles P4 to moles Cl2.
from step (1), 1 mole P4 reacts with 10 moles Cl2
so ...
3.672 moles P4 x (10 moles Cl2 / 1 mole P4) = 36.72 moles Cl2
**** 5. convert moles back to mass ****
from periodic table, atomic mass Cl = 35.45 g/mole.. so Cl2 = 70.90 g/mole
36.72 moles Cl2 x (70.90 g/mole) = 2603 g
since 455 g has 3 sig figs, and since this was all multiplication and division, your final answer needs to have only 3 sig figs...
mass Cl2 = 2.60 kg
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