芜湖弋矶山医院肿瘤科:一道化学题,大虾们请进!
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Na2CO3 + 2HCl = 2NaCl + H2O + CO2(↑)
106g 73g 44g
Xg Yg 0.44g
根据化学方程式得:
m(Na2CO3)=X=1.06g
m1(HCL)=Y=0.73g
又有:m总(HCL) = 60 g * 7.3% = 4.38g
于是:m2(HCl) = m总(HCL) - m1(HCL)
= 4.38 - 0.73 = 3.65g
NaOH + HCL = NaCL + H2O
40g 36.5g
Zg 3.65g
于是:m(NaOH) = Z = 4g
NaOH的质量分数为:
m(NaOH) / [ m(NaOH) + m(Na2CO3) ]
=4/5.06 = 79.1%
Na2CO3 + 2HCl = 2NaCl + H2O + CO2(↑)
106g 73g 44g
Xg Yg 0.44g
根据化学方程式得:
m(Na2CO3)=X=1.06g
m1(HCL)=Y=0.73g
又有:m总(HCL) = 60 g * 7.3% = 4.38g
于是:m2(HCl) = m总(HCL) - m1(HCL)
= 4.38 - 0.73 = 3.65g
NaOH + HCL = NaCL + H2O
40g 36.5g
Zg 3.65g
于是:m(NaOH) = Z = 4g
NaOH的质量分数为:
m(NaOH) / [ m(NaOH) + m(Na2CO3) ]
=4/5.06 = 79.1%
m(HCL)60x7.3%=4.38g
2HCL+Na2CO3=2NaCl+CO2+H2O
73 106 44
m(HCl) m(Na2CO3) 0.44
m(HCl)=0.73g
m(Na2CO3)=1.06g
NaOH+HCl=H2O+NaCl
40 36.5
m(NaOH) (4.38-0.73)g
m(NaOH)=4
NaOH的质量分数是4/(4+1.06)x100%=79.1%
支持楼上的~
m(HCL)60x7.3%=4.38g
2HCL+Na2CO3=2NaCl+CO2+H2O
73 106 44
m(HCl) m(Na2CO3) 0.44
m(HCl)=0.73g
m(Na2CO3)=1.06g
NaOH+HCl=H2O+NaCl
40 36.5
m(NaOH) (4.38-0.73)g
m(NaOH)=4
NaOH的质量分数是4/(4+1.06)x100%=79.1%
m(HCL)60x7.3%=4.38g
2HCL+Na2CO3=2NaCl+CO2+H2O
73 106 44
m(HCl) m(Na2CO3) 0.44
m(HCl)=0.73g
m(Na2CO3)=1.06g
NaOH+HCl=H2O+NaCl
40 36.5
m(NaOH) (4.38-0.73)g
m(NaOH)=4
NaOH的质量分数是4/(4+1.06)x100%=79.1%
同意楼上