If 15 g of #C_2H_6# react with 45 g of #O_2#, how many grams of water will be produced?

Answer 1

#27"g"H_2O#

Now let's write this combustion reaction's chemical equation:

#C_2H_6"(g)" + 7/2O_2"(g)" rarr 2CO_2"(g)" + 3H_2O"(g)"#

We must identify which reactant is limiting since we are given the amounts of multiple reactants. To do this, convert each reactant's molar masses to moles and divide the result by the coefficient in the equation:

#15cancel("g"C_2H_6)((1"mol"C_2H_6)/(30.08cancel("g"C_2H_6))) = 0.50"mol"C_2H_6#
#75cancel("g"O_2)((1"mol"O_2)/(32.00cancel("g" O_2))) = (2.344"mol")/(7/2"(coefficient)") = 0.67"mol"O_2#
Since #C_2H_6# is present in relative deficiency, it is the limiting reactant.
Now, we'll use the stoichiometric relationships (the coefficients) to find the moles of #H_2O# and then water's molar mass to calculate the mass formed.
#0.50cancel("mol"C_2H_6)((3cancel("mol"H_2O))/(1cancel("mol"C_2H_6))((18.02"g"H_2O)/(1cancel("mol"H_2O))) = color(red)(27"g"H_2O#
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Answer 2

To determine the amount of water produced, we first need to balance the chemical equation for the reaction between ethane (C2H6) and oxygen (O2) to form water (H2O) and carbon dioxide (CO2):

2C2H6 + 7O2 -> 4CO2 + 6H2O

Now, let's find the limiting reactant by calculating the moles of each reactant:

Moles of C2H6 = 15 g / molar mass of C2H6 Moles of O2 = 45 g / molar mass of O2

Then, we'll use the stoichiometry of the balanced equation to determine the moles of water produced from the limiting reactant. Finally, we'll convert moles of water to grams:

Moles of water = moles of limiting reactant × (6 moles of H2O / 2 moles of C2H6)

Now, multiply the moles of water by the molar mass of water to find the grams of water produced.

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Answer 3

To find out how many grams of water will be produced when 15 g of ( C_2H_6 ) reacts with 45 g of ( O_2 ), we need to first determine the limiting reactant by using stoichiometry.

  1. Write the balanced chemical equation for the reaction: [ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O ]

  2. Calculate the molar masses of the reactants and products:

  • ( Molar , mass , of , C_2H_6 = 2 \times 12.01 , g/mol + 6 \times 1.01 , g/mol = 30.07 , g/mol )
  • ( Molar , mass , of , O_2 = 2 \times 16.00 , g/mol = 32.00 , g/mol )
  • ( Molar , mass , of , H_2O = 2 \times 1.01 , g/mol + 16.00 , g/mol = 18.02 , g/mol )
  1. Calculate the number of moles of each reactant:
  • Moles of ( C_2H_6 = \frac{15 , g}{30.07 , g/mol} )
  • Moles of ( O_2 = \frac{45 , g}{32.00 , g/mol} )
  1. Determine the limiting reactant:
  • Compare the moles of ( C_2H_6 ) and ( O_2 ) to see which one is present in the lower amount.
  1. Once you have identified the limiting reactant, use stoichiometry to find the moles of water produced from the limiting reactant.

  2. Finally, convert the moles of water to grams using its molar mass.

By following these steps, you can calculate the grams of water produced in the reaction.

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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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