If 15 g of #C_2H_6# react with 45 g of #O_2#, how many grams of water will be produced?
Now let's write this combustion reaction's chemical equation:
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:
By signing up, you agree to our Terms of Service and Privacy Policy
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.
By signing up, you agree to our Terms of Service and Privacy Policy
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.
-
Write the balanced chemical equation for the reaction: [ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O ]
-
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 )
- 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} )
- Determine the limiting reactant:
- Compare the moles of ( C_2H_6 ) and ( O_2 ) to see which one is present in the lower amount.
-
Once you have identified the limiting reactant, use stoichiometry to find the moles of water produced from the limiting reactant.
-
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.
By signing up, you agree to our Terms of Service and Privacy Policy
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.
- How do we make a #5.0*L# volume of aqueous ammonia of #0.70*mol*L^-1# concentration, from conc. ammonia of #14.8*mol*L^-1# concentration?
- How many moles of oxygen are produced by the decomposition of six moles of potassium chlorate in the reaction #2KClO_3 -> 2KCl + 3O_2#?
- A #6*g# mass of iron reacts with a #1.6*g# mass of dioxygen to form #FeO#. What is the reagent in excess, and what is the mass of product formed?
- How many moles of sodium hydroxide will be produced from the complete reaction of 15.0 g sodium in the reaction #2Na + 2H_2O -> 2NaOH + H_2#?
- In the reaction #Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3#, how many moles of #Al(OH)_3# can be made with 2.3 moles of #NaOH# and excess #Al_2(SO_4)3#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7