Question

due on or before sept 8th (monday) homework problems late homework = zero!!!!! use factor - label (dimensional analysis) method to solve the following problems. show your work! 1. (a) how many moles is 1.20 x 10²⁵ 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. To find moles from molecules, use the formula: \( \text{Moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} \)
\[
\text{Moles of } NH_3 = \frac{1.20 \times 10^{25} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}}
\]

Step2: Calculate the moles

\[
\frac{1.20 \times 10^{25}}{6.022 \times 10^{23}} \approx \frac{120 \times 10^{23}}{6.022 \times 10^{23}} \approx \frac{120}{6.022} \approx 19.9 \text{ mol}
\]

Step1: Find molar mass of \( NH_3 \)

Molar mass of \( N = 14.01 \text{ g/mol} \), molar mass of \( H = 1.008 \text{ g/mol} \). For \( NH_3 \), molar mass \( = 14.01 + 3 \times 1.008 = 14.01 + 3.024 = 17.034 \text{ g/mol} \)

Step2: Use moles to find mass

From part (a), moles of \( NH_3 = 19.9 \text{ mol} \). Mass \( = \text{Moles} \times \text{Molar mass} \)
\[
\text{Mass of } NH_3 = 19.9 \text{ mol} \times 17.034 \text{ g/mol} \approx 339 \text{ g (close to 338 g due to rounding)}
\]

Answer:

\( 19.9 \text{ mol} \)

Part (b)