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
Answer: If the theoretical yield is 2.35g then the percent yield is 80.4%
Explanation:
Given: Actual yield = 1.89 g
Theoretical yield = 2.35 g
Formula used to calculate the percentage yield is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that if the theoretical yield is 2.35g then the percent yield is 80.4%
Cool little flowcharts I found off Google Images!
The mass of NaCl formed is 8.307 grams
<u><em> calculation</em></u>
step 1: write the equation for reaction
Na₂CO₃ + 2HCl → 2 NaCl +CO₂ +H₂O
Step 2: find the moles of Na₂CO₃
moles = mass/molar mass
The molar mass of Na₂CO₃ is = (23 x2) + 12 + ( 16 x3) = 106 g/mol
moles = 7.5 g/106 g/mol =0.071 moles
Step 3: use the mole ratio to determine the mole of NaCl
Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl =0.07 x2 =0.142 moles
Step 4: calculate mass of NaCl
mass= moles x molar mass
the molar mass of NaCl= 23 +35.5 =58.5 g/mol
mass = 0.142 moles x 58.5 g/mol =8.307 grams
The name given to these electrons are that they are valence electrons or binding electrons as these are directly involved in chemical Bonding and allow for different compounds to be made.
