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floppykiddo
@floppykiddo
July 2022
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В 200 г 2% раствора NaOH влили 100 г 4% раствора HCl. Найти массу полученного NaCl!!
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Alexei78
Дано
m(ppa NaOH)=200 g
W(NaOH)=2%
m(ppa HCL)=100 g
W(HCL)=4%
---------------------------
m(NaCL)-?
m(NaOH)=200*2%/100%=4 g
m(HCL)=100*4%/100% = 4g
M(NaOH)=40 g/mol
n(NaOH)=m/M=4/40=0.1 mol
M(HCL)=36.5 g/mol
n(HCL)=m/M=4/36.5=0.11 mol
n(NaOH)<n(HCL)
NaOH+HCL-->NaCL+H2O
n(NaOH)=n(NaCL)=0.1 mol
M(NaCL)=58.5 g/mol
m(NaCL)=n*M=0.1 * 58.5 = 5.85 g
ответ 5.85 г
2 votes
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Answers & Comments
m(ppa NaOH)=200 g
W(NaOH)=2%
m(ppa HCL)=100 g
W(HCL)=4%
---------------------------
m(NaCL)-?
m(NaOH)=200*2%/100%=4 g
m(HCL)=100*4%/100% = 4g
M(NaOH)=40 g/mol
n(NaOH)=m/M=4/40=0.1 mol
M(HCL)=36.5 g/mol
n(HCL)=m/M=4/36.5=0.11 mol
n(NaOH)<n(HCL)
NaOH+HCL-->NaCL+H2O
n(NaOH)=n(NaCL)=0.1 mol
M(NaCL)=58.5 g/mol
m(NaCL)=n*M=0.1 * 58.5 = 5.85 g
ответ 5.85 г