Answer:
The answer to your question is 3 moles of AlCl₃
Explanation:
Process
1.- Write and balance the equation
Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃
2.- Determine the limiting reactant
Theoretical proportion 1 mol Al(NO₃)₃ : 3 moles of NaCl
Experimental proportion 4 moles Al(NO₃)₃ : 9 moles NaCl
From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃ increases four times.
3.- Determine the amount of AlCl₃ using proportions
3 moles of NaCl --------------- 1 mol of AlCl₃
9 moles of NaCl ---------------- x
x = (9 x 1) / 3
x = 9 /3
x = 3 moles
Step 1: Write the unbalanced equation,
C₂H₆ + O₂ → CO₂ + H₂<span>O
There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,
C₂H₆ + O₂ → 2 CO₂ + H₂O
Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.
C₂H₆ + O₂ → 2 CO₂ + 3 H₂O
At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.
C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O
Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.
( C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O ) × 2
2 C₂H₆ + 3 O₂ → 4 CO₂ + 6 H₂O
To convert from grams to moles, you divide the number of grams by the molar mass To convert from moles to grams, you multiply by the molar mass. You must first calculate the molar mass of the substance
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