Question

1. the following two reactions are important in the blast furnace production of iron metal from iron ore (fe₂o₃):

2c(s) + o₂(g) → 2co(g)
fe₂o₃(s) + 3co(g) → 2fe(s) + 3co₂(g)
using these balanced reactions, how many moles of o₂ are required for the production of 3.19 kg of fe?
a) 42.8 moles
b) 19.0 moles
c) 171 moles
d) 57.1 moles
e) 2.39 moles

1. ammonia can be made by reaction of water with magnesium nitride as shown by the following unbalanced equation:

mg₃n₂(s) + h₂o(l) → mg(oh)₂(s) + nh₃(g)
if this process is 80% efficient, what mass of ammonia can be prepared from 19.0 kg magnesium nitride?
a) 2.6 kg nh3
b) 6.4 kg nh3
c) 5.1 kg nh3
d) 3.2 kg nh3
e) 15 kg nh3

1. iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. in the blood of an adult human, there are approximately 2.64 × 10¹³ red blood cells with a total of 2.90 g of iron. on the average, how many iron atoms are present in each red blood cell? (molar mass fe = 55.85 g/mol)

a) 8.44 × 10⁻¹⁰
b) 1.18 × 10⁹
c) 3.13 × 10²²
d) 2.64 × 10¹³
e) 6.14 × 10⁻²

Explanation:

Question 5

Step1: Convert mass of Fe to moles

Molar mass of Fe is 55.85 g/mol. Mass of Fe is 3.19 kg = 3190 g. Moles of Fe = $\frac{3190\ g}{55.85\ g/mol} \approx 57.12\ mol$.

Step2: Relate Fe moles to CO moles

From the reaction $\ce{Fe_{2}O_{3}(s) + 3CO(g) -> 2Fe(s) + 3CO_{2}(g)}$, 2 moles of Fe are produced from 3 moles of CO. So moles of CO = $\frac{3}{2} \times$ moles of Fe = $\frac{3}{2} \times 57.12\ mol = 85.68\ mol$.

Step3: Relate CO moles to O₂ moles

From the reaction $\ce{2C(s) + O_{2}(g) -> 2CO(g)}$, 2 moles of CO are produced from 1 mole of O₂. So moles of O₂ = $\frac{1}{2} \times$ moles of CO = $\frac{1}{2} \times 85.68\ mol = 42.84\ mol \approx 42.8\ mol$.

Step1: Balance the chemical equation

The balanced equation is $\ce{Mg_{3}N_{2}(s) + 6H_{2}O(l) -> 3Mg(OH)_{2}(s) + 2NH_{3}(g)}$.

Step2: Calculate moles of $\ce{Mg_{3}N_{2}}$

Molar mass of $\ce{Mg_{3}N_{2}}$: (3×24.31) + (2×14.01) = 72.93 + 28.02 = 100.95 g/mol. Mass of $\ce{Mg_{3}N_{2}}$ is 19.0 kg = 19000 g. Moles of $\ce{Mg_{3}N_{2}}$ = $\frac{19000\ g}{100.95\ g/mol} \approx 188.2\ mol$.

Step3: Calculate theoretical moles of $\ce{NH_{3}}$

From the balanced equation, 1 mole of $\ce{Mg_{3}N_{2}}$ produces 2 moles of $\ce{NH_{3}}$. So theoretical moles of $\ce{NH_{3}}$ = 2×188.2\ mol = 376.4\ mol.

Step4: Account for efficiency

Efficiency is 80%, so actual moles of $\ce{NH_{3}}$ = 0.8×376.4\ mol = 301.12\ mol.

Step5: Calculate mass of $\ce{NH_{3}}$

Molar mass of $\ce{NH_{3}}$ is 17.03 g/mol. Mass of $\ce{NH_{3}}$ = 301.12\ mol × 17.03\ g/mol ≈ 5128\ g = 5.13\ kg ≈ 5.1\ kg.

Step1: Calculate moles of Fe

Molar mass of Fe is 55.85 g/mol. Mass of Fe is 2.90 g. Moles of Fe = $\frac{2.90\ g}{55.85\ g/mol} \approx 0.052\ mol$.

Step2: Calculate number of Fe atoms

Number of atoms = moles × Avogadro's number ($6.022×10^{23}$ atoms/mol). Number of Fe atoms = 0.052\ mol × $6.022×10^{23}$ atoms/mol ≈ $3.13×10^{22}$ atoms.

Step3: Calculate atoms per red blood cell

Number of red blood cells is $2.64×10^{13}$. Atoms per red blood cell = $\frac{3.13×10^{22}}{2.64×10^{13}} \approx 1.18×10^{9}$ atoms.

Answer:

A) 42.8 moles

Question 6