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 of Mg uses atmospheric $\text{O}_2$ (unlimited supply). Mg is the limiting reactant, $\text{O}_2$ is excess because $\text{O}_2$ is abundant in air, so Mg will be fully consumed first.

Step3: Balance given chemical equation

Match atom counts for each element:

1. Balance Ca: 1 on left, so 1 for $\text{CaCO}_3$
2. Balance Cl: 2 in $\text{CaCl}_2$, so 2 for $\text{NaCl}$
3. Balance Na: 2 in $\text{NaCl}$, so 2 for $\text{NaHCO}_3$
4. Verify C, H, O: 2 C from 2 $\text{NaHCO}_3$ matches 1 $\text{CaCO}_3$ + 1 $\text{CO}_2$; 2 H matches 1 $\text{H}_2\text{O}$; 6 O matches 3+2+1 O.

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

Step4: Calculate molar masses

Use atomic masses: Mg=24.305 g/mol, O=16.00 g/mol, Ca=40.08 g/mol, Cl=35.45 g/mol, Na=22.99 g/mol, C=12.01 g/mol, H=1.008 g/mol

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

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

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

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

Molecules = moles $\times$ Avogadro's number ($6.022\times10^{23}$ 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: $\text{Mg}$; Excess reactant: $\text{O}_2$. Explanation: Atmospheric oxygen is infinitely abundant compared to the finite amount of magnesium used, so magnesium will be fully consumed first.
2. $\text{CaCl}_2 + 2\text{NaHCO}_3

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

1. Molar masses:

• $\text{Mg}$: 24.305 g/mol
• $\text{O}_2$: 32.00 g/mol
• $\text{MgO}$: 40.305 g/mol
• $\text{CaCl}_2$: 110.98 g/mol
• $\text{Na}_2\text{CO}_3$: 105.99 g/mol
• $\text{CaCO}_3$: 100.09 g/mol
• $\text{NaCl}$: 58.44 g/mol
• $\text{CO}_2$: 44.01 g/mol
• $\text{H}_2\text{O}$: 18.016 g/mol

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