How many grams of O_2(g) are needed to react completely with 3.95 g of H_2(g) according to the following balanced chemical equation: 2H_2(g) + O_2(g)--> 2H_2_O(g)
Molarity (M) is the number of moles of a solute per 1 L of solution--> M= (mol solute)/(L solution). How many moles of permanganate are in 8.23 L of a 3.78 M solution?
@bobpursley
Molarity= 3.78/8.23= 0.459
0.459=molarity
0.459 x 8.23= 3.78moles
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I got 31.6 g of O2 for #1 is that correct?
To calculate the number of grams of O2 that are needed to react completely with 3.95 g of H2, you need to use stoichiometry.
Given the balanced chemical equation: 2H2(g) + O2(g) -> 2H2O(g)
First, calculate the number of moles of H2 using its molar mass. The molar mass of H2 is 2 g/mol.
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 3.95 g / 2 g/mol
moles of H2 = 1.975 mol
According to the balanced equation, the mole ratio between H2 and O2 is 2:1. This means that for every 2 moles of H2, 1 mole of O2 is needed.
moles of O2 = 1.975 mol H2 * (1 mol O2 / 2 mol H2)
moles of O2 = 0.9875 mol
Finally, calculate the mass of O2 using its molar mass. The molar mass of O2 is 32 g/mol.
mass of O2 = moles of O2 * molar mass of O2
mass of O2 = 0.9875 mol * 32 g/mol
mass of O2 = 31.6 g
So you are correct. The number of grams of O2 needed to react completely with 3.95 g of H2 is 31.6 g.
Regarding the second question, you are given that the molarity (M) is 3.78 M, and the volume (V) is 8.23 L.
Molarity is defined as moles of solute divided by liters of solution.
So, moles of solute = molarity * volume
moles of solute = 3.78 M * 8.23 L
moles of solute = 31.1714 mol
Therefore, there are 31.1714 moles of permanganate in the 8.23 L of a 3.78 M solution.
I hope this helps! Let me know if you have any further questions.
For question #1, to determine how many grams of O2(g) are needed, you need to use stoichiometry. Lets calculate it step by step:
1. Determine the molar mass of H2(g) and O2(g):
Molar mass of H2 = 2.02 g/mol
Molar mass of O2 = 32.00 g/mol
2. Convert the given mass of H2(g) to moles:
moles of H2 = mass of H2 / molar mass of H2
= 3.95 g / 2.02 g/mol
= 1.95 mol H2
3. Use the balanced equation to determine the mole ratio between H2 and O2:
From the balanced equation, 2 moles of H2 react with 1 mole of O2.
Therefore, moles of O2 required = 1.95 mol H2 * (1 mol O2 / 2 mol H2)
= 0.975 mol O2
4. Convert the moles of O2 to grams:
grams of O2 = moles of O2 * molar mass of O2
= 0.975 mol O2 * 32.00 g/mol
= 31.2 g O2
Therefore, the correct answer for question #1 is 31.2 g of O2.