Question

stoichiometry quiz

1. solid potassium reacts with chlorine gas to produce potassium chloride. what mass of potassium needs to react if 10.48 g of potassium chloride is formed? (4 points)

2k$_{(s)}$ + cl$_{2(g)}$ → 2kcl$_{(s)}$

1. how many moles of co$_{2}$ are needed to react when do 2.41 x10$^{24}$ molecules of o$_{2}$ form? (3 points)

co$_{2(g)}$ → o$_{2(g)}$ + c (g)

1. propane (c$_{3}$h$_{8}$) gas burns with oxygen gas. if 3.78 x10$^{23}$ molecules of water are produced, how many grams of propane reacted? balance the reaction below. (5 points)

□c$_{3}$h$_{8(g)}$ + □o$_{2(g)}$ → □co$_{2(g)}$ + □ h$_{2}$o$_{(g)}$

Explanation:

Question 1

Step 1: Calculate moles of KCl

Molar mass of \( KCl \): \( M(K) = 39.10 \, g/mol \), \( M(Cl) = 35.45 \, g/mol \), so \( M(KCl) = 39.10 + 35.45 = 74.55 \, g/mol \).
Moles of \( KCl = \frac{mass}{molar \, mass} = \frac{10.48 \, g}{74.55 \, g/mol} \approx 0.1406 \, mol \).

Step 2: Relate moles of K to KCl (from balanced equation \( 2K + Cl_2

ightarrow 2KCl \))
Mole ratio of \( K : KCl = 2 : 2 = 1 : 1 \). So moles of \( K = \) moles of \( KCl = 0.1406 \, mol \).

Step 3: Calculate mass of K

Molar mass of \( K = 39.10 \, g/mol \).
Mass of \( K = moles \times molar \, mass = 0.1406 \, mol \times 39.10 \, g/mol \approx 5.498 \, g \).

Step 1: Balance the equation (given: \( CO_{2(g)}

ightarrow O_{2(g)} + C_{(g)} \))
Balanced equation: \( 2CO_{2(g)}
ightarrow O_{2(g)} + 2C_{(g)} \) (mole ratio \( CO_2 : O_2 = 2 : 1 \)).

Step 2: Calculate moles of \( O_2 \)

Using Avogadro’s number (\( N_A = 6.022 \times 10^{23} \, molecules/mol \)):
Moles of \( O_2 = \frac{number \, of \, molecules}{N_A} = \frac{2.41 \times 10^{24} \, molecules}{6.022 \times 10^{23} \, molecules/mol} \approx 4.00 \, mol \).

Step 3: Relate moles of \( CO_2 \) to \( O_2 \) (from balanced equation)

Mole ratio \( CO_2 : O_2 = 2 : 1 \), so moles of \( CO_2 = 2 \times \) moles of \( O_2 = 2 \times 4.00 \, mol = 8.00 \, mol \).

Step 1: Balance the combustion reaction of propane (\( C_3H_8 + O_2

ightarrow CO_2 + H_2O \))

• Balance C: \( C_3H_8

ightarrow 3CO_2 \) (3 C on left, 3 on right).

• Balance H: \( C_3H_8

ightarrow 4H_2O \) (8 H on left, 8 on right: \( 4 \times 2 = 8 \)).

• Balance O: \( 3CO_2 + 4H_2O \) has \( 3 \times 2 + 4 \times 1 = 10 \) O. So \( O_2 \) needs \( \frac{10}{2} = 5 \).

Balanced equation: \( \boldsymbol{1}C_3H_8 + \boldsymbol{5}O_2
ightarrow \boldsymbol{3}CO_2 + \boldsymbol{4}H_2O \).

Step 2: Calculate moles of \( H_2O \)

Molecules of \( H_2O = 3.78 \times 10^{23} \), so moles of \( H_2O = \frac{3.78 \times 10^{23}}{6.022 \times 10^{23} \, mol^{-1}} \approx 0.6277 \, mol \).

Step 3: Relate moles of \( C_3H_8 \) to \( H_2O \) (mole ratio \( C_3H_8 : H_2O = 1 : 4 \))

Moles of \( C_3H_8 = \frac{moles \, of \, H_2O}{4} = \frac{0.6277 \, mol}{4} \approx 0.1569 \, mol \).

Step 4: Calculate mass of \( C_3H_8 \)

Molar mass of \( C_3H_8 \): \( 3(12.01) + 8(1.008) = 36.03 + 8.064 = 44.094 \, g/mol \).
Mass of \( C_3H_8 = 0.1569 \, mol \times 44.094 \, g/mol \approx 6.92 \, g \).

Answer:

Approximately \( 5.50 \, g \) (or \( 5.498 \, g \))

Question 2