将亚硫酸钠与硫化钠的混合物若干克配成50克溶液,
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因为Na2SO3+2HCl=2NaCl+SO2↑+H2O ① ,

Na2S+2HCl=2NaCl+H2S↑②,

SO2+2H2S=3S↓+2H2O ,

因此,放出的气体是SO2或H2S.

根据质量守恒定律可知,m(S)+m(气体)=(50g+50g)-96.9g=3.1g,

(1)当放出的气体是SO2时,n1(SO2)=0.448L/(22.4L/mol)=0.02mol,

则m1(SO2)=0.02mol*64g/mol=1.28g,

所以m(S)=3.1g-1.28g=1.82g,

SO2 + 2H2S = 3S↓ + 2H2O

1mol~2mol~3*32g

n2(SO2)~n(H2S)~1.82g

列比例式解得:n2(SO2)=0.019mol,n(H2S)=0.038mol

故n(SO2)=n1(SO2)+n2(SO2)=0.039mol

根据化学方程式①知,n(Na2SO3)=n(SO2)=0.039mol,

根据化学方程式②知,n(Na2S)=n(H2S)=0.038mol.

(2)当放出的气体是H2S时,n1(H2S)=0.448L/(22.4L/mol)=0.02mol,

则m1(H2S)=0.02mol*34g/mol=0.68g,

所以m(S)=3.1g-0.68g=2.42g,

SO2 + 2H2S = 3S↓ + 2H2O

1mol~2mol~3*32g

n(SO2)~n2(H2S)~2.42g

列比例式解得:n(SO2)=0.0252mol,n2(H2S)=0.0504mol

故n(H2S)=n1(H2S)+n2(H2S)=0.0704mol

根据化学方程式①知,n(Na2SO3)=n(SO2)=0.0252mol,

根据化学方程式②知,n(Na2S)=n(H2S)=0.0704mol.

答:混合物中亚硫酸钠与硫化钠的物质的量为0.039mol、0.038mol,或混合物中亚硫酸钠与硫化钠的物质的量为0.0252mol、0.0704mol.