Question

Determine the number of moles and mass requested for each reaction in Exercise 4.42.
Refer to Exercise 4.42.
Write the balanced equation, then outline the steps necessary determine the information equation in each of the following:
(a) the number of moles than the mass of the chlorine Cl2 required to react with 10.0 g of sodium metal Na to produce sodium chloride NaCl
(b) the number of moles and the mass of the oxygen formed by the decomposition of 1.252 gram of Mercury (II) oxide
(c)the number of the moles and the mass of the sodium nitrate NaNO3 required to produce 128 gram of Oxygen (NaNO2 is the other product)
(d)the number of moles and the mass of the carbon dioxide formed by the combustion of 20.0 kg of carbon in excess of oxygen
(e)the number of moles and the mass of the copper(II) carbonate needed to produce 1.500 kg of copper II oxide (CO2 is the other product)

Answer 1

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

1. Step 1: Write the balanced equation
2. Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
3. Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. 46g of Na react with 1 mol of Cl₂. Since the molar mass of Cl₂ is 71g/mol, then 46g of Na react with 71g of Cl₂.

Step 3:

$10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}$

$10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}$

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

216.5g of HgO form 0.5 moles of O₂. 216.5g of HgO form 16g of O₂.

Step 3:

$1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}$

$1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}$

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

16g of O₂ come from 1 mol of NaNO₃. 16g of O₂ come from 85g of NaNO₃.

Step 3:

$128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}$

$128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}$

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

12 g of C form 1 mol of CO₂. 12 g of C form 44g of CO₂.

Step 3:

$20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2$

$[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2$[/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

79.5g of CuO come from 1 mol of CuCO₃. 79.5g of CuO come from 123.5g of CuCO₃.

Step 3:

$1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}$