Answer: option d
Explanation:
Project teams must be proactive. Each member of the team should empowered and each member can share ideas and experience in order to make the WBS more precise.
Answer:
92.4 grams.
Explanation:
- From the balanced reaction:
<em>CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,</em>
1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.
- We need to calculate the no. of moles of (104 g) of CaCO₃:
<em>no. of moles of CaCO₃ = mass/molar mass</em> = (104 g)/(100.08 g/mol) = <em>1.039 mol.</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.
∴ 1.039 mole of CaCO₃ produce → 1.039 mole of CaCl₂.
∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.039 mol)(110.98 g/mol) = 114.3 g.
∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.
Percent yield of the reaction = 80.15%, theoretical yield = 115.3 g.
<em>∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 </em>= [(80.15%)/(115.3 g)] / 100 = <em>92.42 g ≅ 92.4 g.</em>
Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2
