Consider the following balanced chemical equation: 2Ca + O2 (g) -> 2CaO. What mass of CaO, in grams, is formed when 10.0 grams of Ca reacts with 5.00 grams of O2? 10.0 g CaO 17.5 g CaO 8.76 g CaO 5.00 g CaO 14.0 g CaO
The Correct Answer and Explanation is:
To solve this problem, we first need to determine the limiting reactant between calcium (Ca) and oxygen (O2) and then calculate the mass of calcium oxide (CaO) produced.
Step 1: Write the balanced equation
The balanced chemical equation is:
2Ca(s)+O2(g)→2CaO(s)2Ca (s) + O_2 (g) \rightarrow 2CaO (s)2Ca(s)+O2(g)→2CaO(s)
This means that 2 moles of calcium react with 1 mole of oxygen gas to produce 2 moles of calcium oxide.
Step 2: Calculate moles of each reactant
To convert grams of each reactant to moles, we use the molar masses of calcium and oxygen:
- Molar mass of calcium (Ca) = 40.08 g/mol
- Molar mass of oxygen (O2) = 32.00 g/mol
Moles of calcium (Ca):
moles of Ca=10.0 g40.08 g/mol=0.249 mol\text{moles of Ca} = \frac{10.0 \, \text{g}}{40.08 \, \text{g/mol}} = 0.249 \, \text{mol}moles of Ca=40.08g/mol10.0g=0.249mol
Moles of oxygen (O2):
moles of O2=5.00 g32.00 g/mol=0.15625 mol\text{moles of O}_2 = \frac{5.00 \, \text{g}}{32.00 \, \text{g/mol}} = 0.15625 \, \text{mol}moles of O2=32.00g/mol5.00g=0.15625mol
Step 3: Determine the limiting reactant
From the balanced equation, 2 moles of Ca react with 1 mole of O2. Therefore, the mole ratio is:Ca:O2=2:1\text{Ca} : \text{O}_2 = 2 : 1Ca:O2=2:1
For 0.249 moles of calcium, the required moles of oxygen are:moles of O2=0.2492=0.1245 mol\text{moles of O}_2 = \frac{0.249}{2} = 0.1245 \, \text{mol}moles of O2=20.249=0.1245mol
Since we have 0.15625 moles of O2, which is more than enough (0.1245 mol), calcium (Ca) is the limiting reactant.
Step 4: Calculate the moles of CaO produced
From the balanced equation, 2 moles of Ca produce 2 moles of CaO. Therefore, the moles of CaO produced will be equal to the moles of Ca reacted, which is 0.249 moles.
Step 5: Convert moles of CaO to grams
Now, we calculate the mass of CaO produced. The molar mass of CaO is:Molar mass of CaO=40.08 g/mol (Ca)+16.00 g/mol (O)=56.08 g/mol\text{Molar mass of CaO} = 40.08 \, \text{g/mol (Ca)} + 16.00 \, \text{g/mol (O)} = 56.08 \, \text{g/mol}Molar mass of CaO=40.08g/mol (Ca)+16.00g/mol (O)=56.08g/mol
Now, multiply the moles of CaO by its molar mass:mass of CaO=0.249 mol×56.08 g/mol=13.95 g\text{mass of CaO} = 0.249 \, \text{mol} \times 56.08 \, \text{g/mol} = 13.95 \, \text{g}mass of CaO=0.249mol×56.08g/mol=13.95g
Step 6: Round the result
The answer is approximately 14.0 g of calcium oxide (CaO).
Final Answer:
The correct mass of CaO formed is 14.0 g.
