Balance the equation NaOH + Al rightarrow Al_2 O_3 + H_2 + Ha_2 O kMnO_4 rightar

Question

Balance the equation NaOH + Al rightarrow Al_2 O_3 + H_2 + Ha_2 O kMnO_4 rightarrow k_2 O + mnO + O_2 H_2 O + O_2 + A_s rightarrow HA_s O_2 Av_2 O_3 rightarrow Au + O_2 Calculate a The number of molecules in 1.058 mole of H_2 O. b The number of atoms in 0.750 mole of Fe. c The mass of 1.2046 times 10^23 atoms of Cu. Calculate the molar masses (in g/mol) for: a KCl b Na_2 CO_3 Given the reaction represented by the balanced equation: CH_4 (g) + 3Cl_2 (g) rightarrow 3 HCl (g) + CHCl_3 (g) a Calculate the number of grams CHCl_3 that could be produced by 105 g cl_2 with excess CH_4. b If 10.0 g CHCl_3 were actually produced, calculated the yield.

Explanation / Answer

Balanced the equation-

i) 6NaOH + 2Al --------> Al2O3 + 3H2 + 3Na2O

ii) 4 KMnO4 -----------> 2 K2O + 4 MnO + O2

iii) O2 + As + 2 H20    -->    HAsO2

O2 is being reduced and the As is being oxidized.

Here are the two 1/2 reactions:

      O2 + 4 H+ + 4 e- -->    2 H2O

        As + 2 H20 -----> HAsO2 + 3H+ + 3 e-

Multiply the 1/2 reactions so that the number of electrons lost = number of electrons gained and then add the two 1/2 reactions together.

3 O2 + 12 H+ + 12 e - -->    6 H2O

4As + 8 H20 -----> 4HAsO2 + 12H+ + 12 e-

Balanced equation-

3 O2 + 4As + 2 H20    -->    4HAsO2

iv) 4Au + 3O2 = 2Au2O3

2)

a) The number of molecules in 1.058 mole of H2O

In 1mol there are 6.022*10^23 molecules of H2O

So in 1.058 moles there are 6.022*10^23 molecules x1.058

= 6.37*10^23 molecules of H2O

b) The number of atoms in 0.750 mole of Fe

There are 6.022*10^23 atoms in1mole of Fe

So, 0.750 mole of Fe there are 0.750 x 6.02 x 10^23 atoms

= 4.52 x 10^23 atoms Fe

c)

The mass of 1.2046 x 10^23 atoms of cu

1 mole of copper will contain 6.022*10^23 atoms of copper.

Copper's molar mass is equal to 63.546 g.

In other words,

6.022*10^23 atoms of copper have a mass of 63.456 g.

1.2046 x 10^23 atoms of copper have a mass of 63.456 g x (1.2046 x 10^23 atoms /6.022*10^23)

= 12.693 g

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