Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows:
Zn(s) + 2HCl(aq) -----> ZnCl(subscript 2)(aq) + H(sub. 2)(g)
( Cu does not react with HCl.) When 0.5065g of a certain brass alloy is reacted with excess Hcl, 0.0985g ZnCl(sub 2) is eventually isolated.
a.) What is the composition of brass by mass?
b.)How could this result be checked without changing the above procedure?
First off, where do I even start for part a? I don't understand how to use what the equation gives me and the given masses to find the percent composition of brass by mass. Help please?? Part b also confuses me..
Zn(s) + 2HCl(aq) -----> ZnCl(subscript 2)(aq) + H(sub. 2)(g)
( Cu does not react with HCl.) When 0.5065g of a certain brass alloy is reacted with excess Hcl, 0.0985g ZnCl(sub 2) is eventually isolated.
a.) What is the composition of brass by mass?
b.)How could this result be checked without changing the above procedure?
First off, where do I even start for part a? I don't understand how to use what the equation gives me and the given masses to find the percent composition of brass by mass. Help please?? Part b also confuses me..
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to answer part A, you have to work backwards.
from the equation
Zn(s) + 2HCl(aq) -----> ZnCl2(aq) + H2(g)
we are given that 0.0985g of ZnCl2 is formed. hence find the number of moles of ZnCl2 produced which is 0.0985 / (65.38 + 2(35.45)) = 0.000722776
Next look at the mole ratio of Zn:ZnCl2 which is 1:1, so 0.000722776 moles of Zn was used in this reaction. Next find the mass of Zn reacted which would be 0.000722776 moles x (65.38g/mol) = 0.047255136g.
Hence in 0.5065g of brass alloy, there is 0.047255136g of Zn in it and 0.459244864g of Cu
Part B. since you can't change the procedure, the other way to find out Zn is to use the other product, H2(g). you can use the gas equation PV=nRT and find out the amount of of H2 gas produced. The reaction could take place in a closed system with a tube attached to the flask to collect the gas somewhere. With the volume of gas obtained, plug it into the equation to get the number of moles of H2(g) n. Then it's just stoichiometry again to find out the mass of Zn
from the equation
Zn(s) + 2HCl(aq) -----> ZnCl2(aq) + H2(g)
we are given that 0.0985g of ZnCl2 is formed. hence find the number of moles of ZnCl2 produced which is 0.0985 / (65.38 + 2(35.45)) = 0.000722776
Next look at the mole ratio of Zn:ZnCl2 which is 1:1, so 0.000722776 moles of Zn was used in this reaction. Next find the mass of Zn reacted which would be 0.000722776 moles x (65.38g/mol) = 0.047255136g.
Hence in 0.5065g of brass alloy, there is 0.047255136g of Zn in it and 0.459244864g of Cu
Part B. since you can't change the procedure, the other way to find out Zn is to use the other product, H2(g). you can use the gas equation PV=nRT and find out the amount of of H2 gas produced. The reaction could take place in a closed system with a tube attached to the flask to collect the gas somewhere. With the volume of gas obtained, plug it into the equation to get the number of moles of H2(g) n. Then it's just stoichiometry again to find out the mass of Zn