Question

question (part banning) - semester 6 / unit 6 - stoichiometry
q19. choose the best answer. (solve the stoichiometry problem using the periodic table. use significant figures. round the atomic mass number to two decimal places.)
how many molecules are in 5.00 g of nacl?
a. 5.1 × 10²³ molecules
b. 5.0 × 10²³ molecules
c. 5.0 × 10²² molecules
d. 5.1 × 10²² molecules
q20. fill in the answer by typing the correct number in the blank. (solve the stoichiometry problem using the periodic table. use significant figures. round the atomic mass number to two decimal places.)
what is the percent composition of sodium and chlorine in nacl?
na: ______%
cl: ______%
(round the percent to two decimal places.)
q21. fill in the answer by typing the correct number in the blank. (solve the stoichiometry problem using the periodic table. use significant figures. round the atomic mass number to two decimal places.)
what is the percent composition of zro₂?
zr: ______%
o: ______%
(round the percent to two decimal places.)

Explanation:

Step1: Calculate NaCl molar mass

Atomic mass of Na = 22.99 g/mol, Cl = 35.45 g/mol.
Molar mass of NaCl = $22.99 + 35.45 = 58.44$ g/mol

Step2: Find moles of NaCl

Moles = $\frac{\text{mass}}{\text{molar mass}} = \frac{5.0}{58.44} \approx 0.08556$ mol

Step3: Calculate number of molecules

Use Avogadro's number $N_A = 6.022 \times 10^{23}$ molecules/mol.
Molecules = $0.08556 \times 6.022 \times 10^{23} \approx 5.15 \times 10^{22}$
Round to 2 significant figures: $5.2 \times 10^{22}$ (closest option is $5.1 \times 10^{22}$)

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Step1: Find NaCl molar mass

Molar mass of NaCl = $22.99 + 35.45 = 58.44$ g/mol

Step2: Calculate % Na in NaCl

$\% \text{Na} = \frac{22.99}{58.44} \times 100 \approx 39.34\%$

Step3: Calculate % Cl in NaCl

$\% \text{Cl} = \frac{35.45}{58.44} \times 100 \approx 60.66\%$

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Step1: Find $Rb_2O$ molar mass

Atomic mass of Rb = 85.47 g/mol, O = 16.00 g/mol.
Molar mass of $Rb_2O = (2 \times 85.47) + 16.00 = 186.94$ g/mol

Step2: Calculate % Rb in $Rb_2O$

$\% \text{Rb} = \frac{2 \times 85.47}{186.94} \times 100 \approx 90.93\%$

Step3: Calculate % O in $Rb_2O$

$\% \text{O} = \frac{16.00}{186.94} \times 100 \approx 9.09\%$

Answer:

1. $\boldsymbol{5.1 \times 10^{22}}$ molecules
2. Na: $\boldsymbol{39.34}$%, Cl: $\boldsymbol{60.66}$%
3. Rb: $\boldsymbol{90.93}$%, O: $\boldsymbol{9.09}$%