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Theoretically the mass is 62,3018 g.
2.43 X 1023 atoms aluminum (1 mole Al/6.022 X 1023)(26.98 grams/1 mole Al)(1 pound/454 grams) = 0.024 pounds of aluminum ==================( about 2/5 of an ounce )
The number of grams of Aluminum in one mole of Aluminum is the number under the element's symbol, on the periodic table of the elements, which is 26.981. With compounds, water, for example, is an oxygen and two hydrogens. So you would add 15.999 to 2(1.0079). You multiply the hydrogen's number by two because there's two of them.
For this you need the atomic mass of Al. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.2.00 moles Al × (27.0 grams) = 54.0 grams Al
174.107g
3 grams
24 / 40 = x / 4.00 x = 2.4 g Mg
Because the atomic mass of aluminum is 26.99, this means that there are 26.99 grams of aluminum in one mole. For 3 moles multiply 26.99 by 3 = 80.97grams.
Also How many grams and what volume of fluorine (@ STP) could be liberated at the anode? Also How many hours would the electrolysis need to continue to produce 75g of aluminum with a current of 15 amperes?
Number of moles = mass of the sample/atomic weight of Al = 96,7/26,987 = 3,58
Theoretically the mass is 62,3018 g.
2.43 X 1023 atoms aluminum (1 mole Al/6.022 X 1023)(26.98 grams/1 mole Al)(1 pound/454 grams) = 0.024 pounds of aluminum ==================( about 2/5 of an ounce )
The number of grams of Aluminum in one mole of Aluminum is the number under the element's symbol, on the periodic table of the elements, which is 26.981. With compounds, water, for example, is an oxygen and two hydrogens. So you would add 15.999 to 2(1.0079). You multiply the hydrogen's number by two because there's two of them.
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For this you need the atomic mass of Al. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.2.00 moles Al × (27.0 grams) = 54.0 grams Al
If the aluminum bubbles it will be a chemical change.
174.107g