Question

late homework = zero use factor-label (dimensional analysis) method to solve the following 1. (a) how many moles is 1.20 × 10²⁸ molecules nh₃ (ammonia)? 19.9 mol nh₃ (b) what mass is this number of molecules? answer: 338 g nh₃

Explanation:

Part (a)

Step1: Recall Avogadro's number

Avogadro's number is \( 6.022 \times 10^{23} \) molecules/mol. We use the conversion factor \( \frac{1\ \text{mol}}{6.022 \times 10^{23}\ \text{molecules}} \).
\[
\text{Moles of } \ce{NH3} = 1.20 \times 10^{25}\ \text{molecules} \times \frac{1\ \text{mol}}{6.022 \times 10^{23}\ \text{molecules}}
\]

Step2: Calculate the moles

\[
\text{Moles of } \ce{NH3} = \frac{1.20 \times 10^{25}}{6.022 \times 10^{23}} \approx 19.9\ \text{mol}
\]

Step1: Find molar mass of \( \ce{NH3} \)

Molar mass of \( \ce{N} = 14.01\ \text{g/mol} \), molar mass of \( \ce{H} = 1.008\ \text{g/mol} \). Molar mass of \( \ce{NH3} = 14.01 + 3\times1.008 = 17.034\ \text{g/mol} \).

Step2: Use moles from part (a) to find mass

Mass = moles × molar mass. Moles from (a) is \( 19.9\ \text{mol} \).
\[
\text{Mass of } \ce{NH3} = 19.9\ \text{mol} \times 17.034\ \text{g/mol} \approx 339\ \text{g}
\]
(Note: The given answer in the image might have a typo, correct calculation gives ~339 g)

Answer:

\( 19.9\ \text{mol} \)

Part (b)