10.25L
3,42 moles of phosphorus trichloride have 469,6686 g.
The answer is 8,33 moles.
The answer is 0,615 moles.
the answer is 4.7 1] Figure out how many moles of Al and O2 2.5g is. 2] Compare the ratio of the moles from A to the 2:3 ratio in Al2O3; do you have more Al proportionately than O2 or vice versa ? This is called 'finding which reagent is limiting".... 3] Take whichever reagent was limiting and find out how many moles of Al2O3 you can get from it. THen find the mass.
The answer is 14,93 moles.
In the given reaction, the stoichiometry is 1:1 for Fe2O3 to Al2O3. So the number of moles of Al2O3 formed will be the same as the number of moles of Fe2O3 originally present.
Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
so you find the ratio of Aluminum to Oxygenin this case it is 2:3and then because the total mole is 2.16for aluminum:2.16/5*2 = 0.864for oxygen:2.16/5*3 = 1.296hope this helped :D
The balanced reaction equation is 4Al + 3O2 -> 2Al2O3. Therefore, 3 moles of O2 reacts with 4 moles of Al to form 2 moles of Al2O3. Since 0.78 mol of O2 is reacted, the number of moles of Al2O3 formed can be calculated using the stoichiometry of the reaction.
To find the number of moles, first calculate the molar mass of sodium nitrate (NaNO3), which is 85 grams/mol. Then, divide the given mass (2.85 grams) by the molar mass to obtain the number of moles present, which is approximately 0.0335 moles.
3,42 moles of phosphorus trichloride have 469,6686 g.
For this you need the atomic (molecular) mass of Al2O3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. Al2O3= 102 grams408 grams Al / (102 grams) = 4.00 moles Al
Remember the Moles Eq'n. moles = mass(g) / Mr The Mr ( Relative molecular mass of Al2O3) is . 2 x Al = 2 x 27 = 54 3 x O = 3 x 16 = 48 54 + 48 = 102 substituting moles(Al2O3) = 6.8 g / 102 moles = 0.0666.... ( recurs to infinity.
Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.
Dinitrogen tetraoxide, or N2O4 has a molar mass of 92.011 grams per mole. This means there are 0.0435 moles present.
Aluminum oxide, Al2O3, is an ionic compound. There are no molecules in it.
The gram molecular mass of Al2O3 is 2(26.982) + 3 (15.999) = 101.96. Therefore, the number of moles of Al2O3 is 291.257/101.96 = 2.857 moles. Each mole contains two moles of aluminum atoms; therefore the number of aluminum atoms in this mass equals 2 X 2.857 X 6.022 X 1023 = 3.4410 X 1024 atoms.