Question

1. $3 f_{2(g)} + 2 alcl_{3(aq)} \

ightarrow 2 alf_{3(aq)} + 3 cl_{2(g)}$
how many moles of fluorine gas are needed to produce $1.5 \times 10^{24}$ molecules of $cl_2$?

1. $(nh_4)_2cr_2o_7 \

ightarrow cr_2o_3 + 4 h_2o + n_2$
how many molecules of $h_2o$ are produced when $7.8 \times 10^{23}$ formula units of $(nh_4)_2cr_2o_7$ are decomposed?

Explanation:

Problem 1

Step 1: Find moles of \( Cl_2 \)

Use Avogadro's number (\( 6.022 \times 10^{23} \) molecules/mol).
Moles of \( Cl_2 = \frac{1.5 \times 10^{24} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 2.49 \text{ mol} \)

Step 2: Use stoichiometry (mole ratio)

From the reaction: \( 3 \text{ mol } F_2
ightarrow 3 \text{ mol } Cl_2 \) (ratio \( 1:1 \)).
Moles of \( F_2 = \) moles of \( Cl_2 \approx 2.5 \text{ mol} \) (or more precisely, \( \frac{1.5 \times 10^{24}}{6.022 \times 10^{23}} \approx 2.49 \approx 2.5 \) mol)

Step 1: Find moles of \( (NH_4)_2Cr_2O_7 \)

Moles \( = \frac{7.8 \times 10^{23} \text{ formula units}}{6.022 \times 10^{23} \text{ formula units/mol}} \approx 1.295 \text{ mol} \)

Step 2: Use stoichiometry (mole ratio)

From the reaction: \( 1 \text{ mol } (NH_4)_2Cr_2O_7
ightarrow 4 \text{ mol } H_2O \).
Moles of \( H_2O = 1.295 \text{ mol} \times 4 \approx 5.18 \text{ mol} \)

Step 3: Find molecules of \( H_2O \)

Molecules \( = 5.18 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 3.12 \times 10^{24} \) molecules

Answer:

Approximately \( 2.5 \) moles (or \( \boldsymbol{\frac{1.5 \times 10^{24}}{6.022 \times 10^{23}} \approx 2.5} \) mol)

Problem 2