Question

1. ammonia can be made by reaction of water with magnesium nitride as shown by the following unbalanced equation: \\( \text{mg}_3\text{n}_2(\text{s}) + \text{h}_2\text{o}(\text{l}) \

ightarrow \text{mg(oh)}_2(\text{s}) + \text{nh}_3(\text{g}) \\) if this process is 80% efficient, what mass of ammonia can be prepared from 19.0 kg magnesium nitride?\
a) 2.6 kg \\( \text{nh}_3 \\)\
b) 6.4 kg \\( \text{nh}_3 \\)\
c) 5.1 kg \\( \text{nh}_3 \\)\
d) 3.2 kg \\( \text{nh}_3 \\)\
e) 15 kg \\( \text{nh}_3 \\)\
\

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 \times 10^{13} \\) 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 \\( \text{fe} = 55.85 \\, \text{g/mol} \\))\

a) \\( 8.44 \times 10^{-10} \\)\
b) \\( 1.18 \times 10^9 \\)\
c) \\( 3.13 \times 10^{22} \\)\
d) \\( 2.64 \times 10^{13} \\)\
e) \\( 6.14 \times 10^{-2} \\)

Explanation:

Step1: Calculate moles of iron

First, find the moles of iron in 2.90 g. Moles = mass / molar mass. So, moles of Fe = $\frac{2.90\ g}{55.85\ g/mol} \approx 0.0519\ mol$.

Step2: Calculate number of iron atoms

Using Avogadro's number ($6.022 \times 10^{23}\ atoms/mol$), number of Fe atoms = $0.0519\ mol \times 6.022 \times 10^{23}\ atoms/mol \approx 3.13 \times 10^{22}\ atoms$.

Step3: Calculate atoms per red blood cell

Total red blood cells = $2.64 \times 10^{13}$. Atoms per RBC = $\frac{3.13 \times 10^{22}\ atoms}{2.64 \times 10^{13}\ RBCs} \approx 1.18 \times 10^{9}\ atoms/RBC$. Wait, no, wait: Wait, the total iron is 2.90 g, total RBCs are $2.64 \times 10^{13}$. Wait, no, my first step was wrong. Wait, the total iron is 2.90 g, molar mass Fe = 55.85 g/mol. Moles of Fe = 2.90 / 55.85 ≈ 0.0519 mol. Atoms of Fe = 0.0519 mol × 6.022e23 atoms/mol ≈ 3.13e22 atoms. Now, divide by number of RBCs: 3.13e22 atoms / 2.64e13 RBCs ≈ 1.18e9 atoms per RBC? Wait, no, that can't be. Wait, no, the problem says "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". So per RBC, mass of Fe is 2.90 g / 2.64e13 RBCs ≈ 1.098e-13 g/RBC. Then moles of Fe per RBC: 1.098e-13 g / 55.85 g/mol ≈ 1.966e-15 mol/RBC. Then atoms: 1.966e-15 mol × 6.022e23 atoms/mol ≈ 1.18e9 atoms/RBC. Wait, but the options have B) 1.18 × 10⁹. Wait, but let's check again. Wait, 2.90 g Fe total. Moles of Fe: 2.90 / 55.85 ≈ 0.0519 mol. Atoms: 0.0519 × 6.022e23 ≈ 3.13e22 atoms. Divide by 2.64e13 RBCs: 3.13e22 / 2.64e13 ≈ 1.18e9 atoms per RBC. So the answer is B.

Wait, no, wait the options for question 7: A) 8.44 × 10⁻¹⁰, B) 1.18 × 10⁹, C) 3.13 × 10²², D) 2.64 × 10¹³, E) 6.14 × 10⁻². So yes, B is 1.18 × 10⁹, which matches our calculation.

Answer:

B. 1.18 × 10⁹