Question

ii. prelab questions (answer questions in your notebook):

1. what is the balanced chemical equation for the combustion of magnesium?
2. in part 1, what do you expect to the be limiting and excess reactants? explain.
3. balance this chemical equation:

_cacl₂ + _nahco3 -> _caco₃ + _nacl + co₂ + h₂o

1. calculate the molar mass of all the reactants and products in this lab (show work)

• mg, o₂, mgo, cacl₂, na₂co₃, caco₃, nacl, co₂, h₂o

1. calculate moles and molecules of 1.23 grams of cacl₂ (show work)

Explanation:

Step1: Balance Mg combustion equation

Magnesium reacts with oxygen to form magnesium oxide. Balance atoms:

$$2\text{Mg} + \text{O}_2 ightarrow 2\text{MgO}$$

Step2: Identify limiting/excess reactants

Combustion uses atmospheric $\text{O}_2$ (unlimited supply).
Magnesium is the limiting reactant (runs out first), $\text{O}_2$ is excess.

Step3: Balance given double displacement equation

Balance Ca, Cl, Na, C, H, O atoms sequentially:

$$\text{CaCl}_2 + 2\text{NaHCO}_3 ightarrow \text{CaCO}_3 + 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}$$

Step4: Calculate molar masses (use atomic masses: Mg=24.305, O=16.00, Ca=40.08, Cl=35.45, Na=22.99, C=12.01, H=1.008)

• $\text{Mg}$: $24.305\ \text{g/mol}$
• $\text{O}_2$: $2\times16.00 = 32.00\ \text{g/mol}$
• $\text{MgO}$: $24.305 + 16.00 = 40.305\ \text{g/mol}$
• $\text{CaCl}_2$: $40.08 + 2\times35.45 = 110.98\ \text{g/mol}$
• $\text{Na}_2\text{CO}_3$: $2\times22.99 + 12.01 + 3\times16.00 = 105.99\ \text{g/mol}$
• $\text{CaCO}_3$: $40.08 + 12.01 + 3\times16.00 = 100.09\ \text{g/mol}$
• $\text{NaCl}$: $22.99 + 35.45 = 58.44\ \text{g/mol}$
• $\text{CO}_2$: $12.01 + 2\times16.00 = 44.01\ \text{g/mol}$
• $\text{H}_2\text{O}$: $2\times1.008 + 16.00 = 18.016\ \text{g/mol}$

Step5: Calculate moles of $\text{CaCl}_2$

Moles = mass / molar mass
$$\text{Moles of CaCl}_2 = \frac{1.23\ \text{g}}{110.98\ \text{g/mol}} \approx 0.0111\ \text{mol}$$

Step6: Calculate molecules of $\text{CaCl}_2$

Molecules = moles $\times$ Avogadro's number ($6.022\times10^{23}\ \text{mol}^{-1}$)
$$\text{Molecules} = 0.0111\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} \approx 6.68\times10^{21}$$

Answer:

1. $2\text{Mg} + \text{O}_2

ightarrow 2\text{MgO}$

1. Limiting reactant: $\text{Mg}$ (it is the finite reactant being burned); Excess reactant: $\text{O}_2$ (unlimited supply from the atmosphere, so it will not be fully consumed).
2. $\text{CaCl}_2 + 2\text{NaHCO}_3

ightarrow \text{CaCO}_3 + 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}$

1. - $\text{Mg}$: $24.305\ \text{g/mol}$

• $\text{O}_2$: $32.00\ \text{g/mol}$
• $\text{MgO}$: $40.305\ \text{g/mol}$
• $\text{CaCl}_2$: $110.98\ \text{g/mol}$
• $\text{Na}_2\text{CO}_3$: $105.99\ \text{g/mol}$
• $\text{CaCO}_3$: $100.09\ \text{g/mol}$
• $\text{NaCl}$: $58.44\ \text{g/mol}$
• $\text{CO}_2$: $44.01\ \text{g/mol}$
• $\text{H}_2\text{O}$: $18.016\ \text{g/mol}$

1. Moles of $\text{CaCl}_2$: $0.0111\ \text{mol}$; Molecules of $\text{CaCl}_2$: $6.68\times10^{21}$