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slavikrds [6]
3 years ago
12

2c2h2(l) 5o2(g) yeild 4co2(g) 2h2o(g) if the acetylene tank contains 37.0 mol of c2h2 and the oxygen tank contains 81.0 mol of o

2, what is the limiting reactant for this reaction?
Chemistry
1 answer:
Murrr4er [49]3 years ago
6 0

2C2H2(l)  +5O2(g)→ 4Co2(g)  + 2H2O

The   limiting  reactant  for reaction above is O2


<u><em>explanation</em></u>

 The limiting reagent is determined using mole ratio  of both reactant.

 that  is the  mole ratio  of  C2H2:O2  which is  2:5 .

 The mole ratio above  implies  that 37.0 mole  of C2H2   needs  37.0 moles  x5/2=92  moles of  O2.


<em>Since the available moles of O2  was 81.0 mole  and 92  moles  are  required to completely  react with  C2H2</em><em> ,</em> <u>O2   is  the limiting reagent.</u>

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6 0
3 years ago
When the reaction shown is correctly balanced, the coefficients are: kclo3 → kcl + o2?
notka56 [123]
From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl 
coefficient 3 in front of oxygen molecule O2

We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
     KClO3 → KCl + O2
       K=1          K=1
       Cl=1         Cl=1
       O=3          O=2

We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen 
atoms on the right side to balance the oxygens:
     2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.

The number of potassium K atoms and chloride Cl atoms have changed as well:
     2KClO3 → KCl + 3O2
       K=2            K=1
       Cl=2          Cl=1
       O=6           O=6

We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side: 
     2KClO3 → 2KCl + 3O2, which is our final balanced equation.
        K=2           K=2
        Cl=2          Cl=2
        O=6           O=6
The potassium, chlorine, and oxygen atoms are now balanced.

5 0
3 years ago
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4 0
3 years ago
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3 years ago
When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac
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<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

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Given the following parameters

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According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

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Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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