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Total 0/0.5 ...In the explosion of a hydrogen-filled balloon, 0.75 g of hydrogen reacted with 6.9 g of oxygen to form how many grams of water vapor? (Water vapor is the only product.)
i already posted this but i don't think people understood. isnt it because the law of conservation of mass that u just add the grams? but then i tried that and it was wrong. please help!
This is a limiting reagent problem and it errs when it states that "0.75 g hydrogen reacts with 6.9 g oxygen" because not all of the oxygen reacts.
Write the equation and balance it.
2H2 + O2 ==> 2H2O
2a. Convert 0.75 g H2 to moles. moles = grams/molar mass.
2b. Convert 6.9 g oxygen to moles the same way.
3a. Using the coefficients in the balanced equation, convert moles hydrogen to moles H2O.
3b. Same procedure, convert moles oxygen to moles H2O.
3c. In limiting reagent problems, the answer for 3a and 3b are not the same; therefore, one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reactant producing that value is the limiting reagent.
4. Using the correct answer from 3c, convert mole H2O to grams H2O. moles x molar mass = grams.
I arrived at an answer of about 6.75 g H2O produced but I estimated here and there. You need to confirm that.
To determine the number of grams of water vapor formed in the explosion of a hydrogen-filled balloon, you need to use the balanced chemical equation for the reaction between hydrogen and oxygen to form water.
The balanced equation for the reaction is: 2H₂ + O₂ → 2H₂O
From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
To solve the problem, follow these steps:
1. Convert the given masses of hydrogen and oxygen to moles:
- Hydrogen: 0.75 g / (2.0159 g/mol) = 0.372 moles
- Oxygen: 6.9 g / (31.9988 g/mol) = 0.216 moles
2. Determine the limiting reactant by comparing the number of moles of hydrogen and oxygen. The reactant that is completely consumed is the limiting reactant. In this case, it is oxygen since the hydrogen-oxygen ratio is 2:1 in the balanced equation. Therefore, 0.216 moles of oxygen is the limiting reactant.
3. Calculate the number of moles of water produced using the stoichiometry of the balanced equation. From the balanced equation, we know that 2 moles of water are produced for every 1 mole of oxygen. Therefore, the number of moles of water formed is 2 * 0.216 moles = 0.432 moles.
4. Convert the moles of water to grams:
- Water: 0.432 moles * (18.0153 g/mol) = 7.79 g
Therefore, 7.79 grams of water vapor is formed in the explosion of the hydrogen-filled balloon.
It's important to note that when applying the law of conservation of mass, you cannot simply add the masses of the reactants. Instead, you need to consider the stoichiometry of the balanced chemical equation to determine the correct ratio of reactants and products.