При взаимодействии предельного одноатомного спирта массой 3, 7 грамм с натрием выделяется водород объемом 0, 56 литров (н.у. Какова относительная молекулярная масса спирта?

дано

m(CnH2n+1OH) = 3.7 g

V(H2) = 0.56 L

M(CnH2n+1OH) - ?

M(CnH2n+1OH) = (14n+18) g/mol

2CnH2n+1OH + 2Na-->2CnH2n+1ONa+H2

  m(CnH2n+1OH)             V(H2)

    =

2* M(CnH2n+1OH)              Vm

 3.7                0.56

   =      

2*(14n+18)       22.4

0.56*2*(14n+18) = 3.7*22.4

15.68 n + 20.16 = 82.88

n = 4

2C4H9OH+2Na-->2C4H9ONa+H2

M(C4H9OH) = 12*4+1*9+16+1 = 74 g/mol

ответ 74 г/моль

Объяснение:

1)

дано

m(AL) = 27 g

m(Al2S3)-?

2Al+3S-->AL2S3

M(Al) = 27 g/mol

n(AL) = m(AL) / M(Al) = 27 / 27 =1 mol

2n(Al) = n(AL2S3)

n(Al2S3) = 1 / 2 = 0.5 mol

M(Al2S3) = 150 g/mol

m(Al2S3) = n*M  =0.5 / 150 = 75 g

ответ 75 г

2)

дано

m(NaBr) = 68 g

m(Br2) =?

2Na+Br2-->2NaBr

M(NaBr) = 103 g/mol

n(NaBr) = m/M = 68 / 103 = 0.66 mol

n(Br2) = 2n(NaBr)

n(Br2) = 0.66 / 2 = 0.33 mol

M(Br2) = 80*2 = 160 g/mol

m(Br) = n*M = 0.33 *160  = 52.8 g

ответ 52.8 г

3)

дано

m(Mg3P2) = 110 g/mol

m(Mg)-?

2P+3Mg-->Mg3P2

M(Mg3P2) = 134 g/mol

n(Mg3P2) = m/M = 110 / 134 = 0.82 mol

n(Mg3P2) = 3n(Mg)

n(Mg) = 3*0.82 = 2.46 mol

M(Mg) = 24 g/mol

m(Mg) = n*M = 2.46 * 24 = 59.04 g

ответ 59.04 г

1)

дано

m(AL) = 27 g

m(Al2S3)-?

2Al+3S-->AL2S3

M(Al) = 27 g/mol

n(AL) = m(AL) / M(Al) = 27 / 27 =1 mol

2n(Al) = n(AL2S3)

n(Al2S3) = 1 / 2 = 0.5 mol

M(Al2S3) = 150 g/mol

m(Al2S3) = n*M  =0.5 / 150 = 75 g

ответ 75 г

2)

дано

m(NaBr) = 68 g

m(Br2) =?

2Na+Br2-->2NaBr

M(NaBr) = 103 g/mol

n(NaBr) = m/M = 68 / 103 = 0.66 mol

n(Br2) = 2n(NaBr)

n(Br2) = 0.66 / 2 = 0.33 mol

M(Br2) = 80*2 = 160 g/mol

m(Br) = n*M = 0.33 *160  = 52.8 g

ответ 52.8 г

3)

дано

m(Mg3P2) = 110 g/mol

m(Mg)-?

2P+3Mg-->Mg3P2

M(Mg3P2) = 134 g/mol

n(Mg3P2) = m/M = 110 / 134 = 0.82 mol

n(Mg3P2) = 3n(Mg)

n(Mg) = 3*0.82 = 2.46 mol

M(Mg) = 24 g/mol

m(Mg) = n*M = 2.46 * 24 = 59.04 g

ответ 59.04 г