<span>the balanced equation for the reaction is as follows
2C</span>₄H₁₀ + 13O₂ ---> 8 CO₂ + 10H₂<span>O
stoichiometry of C</span>₄H₁₀ to O₂ <span>is 2:13
stoichiometry applies to the molar ratio of reactants and products. Avagadros law states that volume of gas is directly proportional to number of moles of gas when pressure and temperature are constant.
Therefore volume ratio of reactants is equal to molar ratio, volume ratio of C</span>₄H₁₀ to O₂<span> is 2:13
2 L of </span>C₄H₁₀ reacts with 13 L of O₂<span>
then 100 L of </span>C₄H₁₀<span> reacts with 13/2 x 100 = 650 L
therefore 650 L of O</span>₂<span> are required </span>
Answer:
41.54 grams of oxygen are required to burn 13.5 g of acetylene
Explanation:
The balanced reaction is:
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- C₂H₂: 2 moles
- O₂: 5 moles
- CO₂: 4 moles
- H₂O: 2 moles
Being the molar mass of the compounds:
- C₂H₂: 26 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- C₂H₂: 2 moles* 26 g/mole= 52 grams
- O₂: 5 moles* 32 g/mole= 160 grams
- CO₂: 4 moles* 44 g/mole= 176 grams
- H₂O: 2 moles* 18 g/mole= 36 grams
You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

mass of oxygen= 41.54 grams
<u><em>41.54 grams of oxygen are required to burn 13.5 g of acetylene</em></u>
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Matter in chemistry is the weight/mass of something
1.706515172 ×10 to the power of 23
Answer:
B
Explanation:
I mean just think about it where else are the particles going to move? The room right?