I think it’s oxygen and carbon
Answer:
Explanation:
<u>1. Word equation:</u>
- <em>mercury(II) oxide → mercury + oxygen </em>
<u>2. Balanced molecular equation:</u>
<u>3. Mole ratio</u>
Write the ratio of the coefficients of the substances that are object of the problem:
<u>4. Calculate the number of moles of O₂(g)</u>
Use the equation for ideal gases:
<u>5. Calculate the number of moles of HgO</u>
<u>6. Convert to mass</u>
- mass = # moles × molar mass
- molar mass of HgO: 216.591g/mol
- mass = 0.315mol × 216.591g/mol = 68.3g
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
Explanation:
A..........is the correct answer
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