Question

Encuentra las masas de los reactivos y los productos en las siguientes ecuaciones:​

Answer 1

Explicación:

1) Al + Fe - - - >FeS (ya balanceada)

Fe : 55.845 g/mol

S : 32.065 g/mol

Fe + S = 87.91

FeS= 87.91

87.91 g/mol - - - > 87.91 g/mol

2) 2Al + 3Br2 - - - > 2AlBr3 (balanceada)

Al : 26.9815g/mol (2) = 53.963 g/mol

Br : 79.904 g/mol (6) = 479.424 g/mol

2Al + 3Br2 = 533.387 g/mol

2AlBr3 = 533.387 g/mol

533.387 g/mol - - - - > 533.387 g/mol

3) 2B + 3Cl2 - - - - - > 2BCl3 (balanceada)

B: 10.811 g/mol (2) = 21.622 g/mol

Cl : 35.453 g/mol (6) = 212.718 g/mol

2b+3Cl2 = 234.34 g/mol

2BCl3 = 234.34 g/mol

234.34 g/mol - - - - - > 234.34 g/mol

4) 2CaO + 2H2O - - - - - > 2Ca(OH)2 (balanceada)

CaO : 56.078 g/mol (2) = 112.156 g/mol

H2O : 18 g/mol (2) = 36 g/mol

2CaO + 2H2O = 148.156 g/mol

2Ca(OH)2= 148.156 g/mol

148.156 g/mol - - - - - - > 148.156 g/mol

5) Na2O + H2O - - - - > 2NaOH (balanceada)

Na2O : 62 g/mol

H2O : 18 g/mol

Na2O + H2O = 80 g/mol

2NaOH = 80 g/mol

80 g/mol - - - - - - > 80 g/mol

6) SO3 + H2O - - - - - > H2SO4 (ya balanceada)

SO3 : 80.065 g/mol

H2O : 18 g/mol

SO3 + H2O = 98.065 g/mol

H2SO4 = 98.065 g/mol

98.065 g/mol - - - - - > 98.065 g/mol

7) N2+3H2 - - - - - > 2NH3 (balanceada)

N2 : 28.0134 g/mol

H : 1.00784 (6) = 6.0468

N2+3H2= 34.0602 g/mol

2NH3=34.0602 g/mol

34.0602 g/mol - - - - - > 34.0602 g/mol