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omeli [17]
3 years ago
15

Write a balanced chemical equation for the complete combustion of C3H7BO3, a gasoline additive. The products of combustion are C

O2(g), H2O(g), and B2O3(s).
Chemistry
1 answer:
nexus9112 [7]3 years ago
7 0

Answer:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

Explanation:

  • For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.

So, the balanced equation:

<em>2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.</em>

It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.

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5 0
3 years ago
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If 2.0 liters of oxygen gas react with excess carbon monoxide at 273 K and 1.00 atm, how many liters of carbon dioxide can be pr
vlada-n [284]
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5 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
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