Questions: What is the mass of 3.367 moles of Ca(OH)2? 249.5 g 4.544 × 10^-2 g 22.007 g 499.0 g 74.10 g

Transcript text: Question 6 1 pts What is the mass of 3.367 moles of $\mathrm{Ca}(\mathrm{OH})_{2}$ ? 249.5 g $4.544 \times 10^{-2} \mathrm{~g}$ 22.007 g 499.0 g 74.10 g

Solution

Solution Steps

Step 1: Determine the Molar Mass of $\mathrm{Ca(OH)_2}$

To find the molar mass of $\mathrm{Ca(OH)_2}$, we need to sum the atomic masses of its constituent elements:

Calcium (Ca): $40.08 \, \text{g/mol}$
Oxygen (O): $16.00 \, \text{g/mol}$
Hydrogen (H): $1.008 \, \text{g/mol}$

The formula for $\mathrm{Ca(OH)_2}$ includes one calcium atom, two oxygen atoms, and two hydrogen atoms. Therefore, the molar mass is calculated as follows:

\[ \text{Molar mass of } \mathrm{Ca(OH)_2} = 40.08 + 2(16.00) + 2(1.008) = 74.096 \, \text{g/mol} \]

Step 2: Calculate the Mass of 3.367 Moles of $\mathrm{Ca(OH)_2}$

Using the molar mass calculated in Step 1, we can find the mass of 3.367 moles of $\mathrm{Ca(OH)_2}$ by multiplying the number of moles by the molar mass:

\[ \text{Mass} = 3.367 \, \text{moles} \times 74.096 \, \text{g/mol} = 249.5 \, \text{g} \]