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?
steps:
how many moles of H2 is 3.95g?
you need for each mole of H2, half as many moles of O2?
how many grams is that?
moles=molarity*liters
@bobpursley
Molarity= 3.78/8.23= 0.459
0.459=molarity
0.459 x 8.23= 3.78moles
I got 31.6 g of O2 for #1 is that correct?
To determine the number of grams of O2(g) needed to react completely with 3.95 g of H2(g), you need to use the given balanced chemical equation and the molar ratios between the reactants.
1. Start by finding the number of moles of H2(g) using the given mass and the molar mass of H2.
- The molar mass of H2 is 2 g/mol, so the number of moles of H2 is:
Moles of H2 = Mass of H2 / Molar mass of H2
= 3.95 g / 2 g/mol
= 1.975 mol
2. Use the molar ratios between H2(g) and O2(g) from the balanced chemical equation to find the number of moles of O2(g) needed.
- According to the balanced chemical equation: 2H2(g) + O2(g) --> 2H2O(g)
- The molar ratio between H2(g) and O2(g) is 2:1. This means that for every 2 moles of H2(g), 1 mole of O2(g) is required.
- So, the number of moles of O2(g) needed can be calculated as:
Moles of O2 = Moles of H2 / Molar ratio (H2:O2)
= 1.975 mol / 2
= 0.9875 mol
3. Finally, calculate the mass of O2(g) needed using the number of moles and the molar mass of O2.
- The molar mass of O2 is 32 g/mol, so the mass of O2 is:
Mass of O2 = Moles of O2 x Molar mass of O2
= 0.9875 mol x 32 g/mol
= 31.6 g (rounded to one decimal place)
Therefore, 31.6 grams of O2(g) are needed to react completely with 3.95 grams of H2(g).
Moving on to the second question:
To determine the number of moles of permanganate (MnO4-) in 8.23 L of a 3.78 M solution, you can use the given molarity (M) and the volume of the solution.
1. Start by using the formula for molarity:
Molarity (M) = Moles of solute / Liters of solution
2. Rearrange the formula to solve for moles of solute:
Moles of solute = Molarity (M) x Liters of solution
3. Substitute the given values into the formula:
Moles of solute = 3.78 M x 8.23 L
= 31.1574 moles (rounded to four decimal places)
Therefore, there are approximately 31.1574 moles of permanganate (MnO4-) in 8.23 L of a 3.78 M solution.