Suppose 9.5 g of gaseous C2H2 reacts with excess O2 according to the reaction below. What is the mass of CO2 produced? C2H2(g) + O2(g) → CO2(g) + H2O(ℓ)

Answer 1
Firstly we need to correct the equation - it isn't balanced as you wrote it. For the complete combustion of ethyne| #C_2H_2 + 2 1/2 O_2 -> 2CO_2 + H_2O#

According to the stoichiometry, one mole of ethyne is needed to produce two moles of carbon dioxide.

Moles of ethyne = mass of ethyne / #M_r# of ethyne = 9.5/26 = 0.365 moles
Moles of #CO _2# = moles of ethyne x 2 = 0.731
Mass of #CO_2# = moles of #CO_2# x #M_r# of #CO_2# = 0.731 x 44 = 32.15g
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Answer 2

To find the mass of CO2 produced, first, calculate the moles of C2H2 using its molar mass. Then, use the stoichiometry of the reaction to determine the moles of CO2 produced. Finally, multiply the moles of CO2 by its molar mass to find the mass. The molar mass of CO2 is 44.01 g/mol.

  1. Calculate moles of C2H2: ( \text{moles} = \frac{\text{mass}}{\text{molar mass}} )
  2. Use stoichiometry to find moles of CO2 produced.
  3. Convert moles of CO2 to mass: ( \text{mass} = \text{moles} \times \text{molar mass of CO2} ).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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