Aluminium hydroxide, Al(OH)3 has molar mass 27.0+3(17.0) = 78.0
Amount of Al(OH)3 = 39.0/78.0 = 0.500mol
There are 0.5 moles of aluminium hydroxide in a 39.0g pure sample.
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
To determine the grams of aluminum hydroxide obtained from 17.2 grams of aluminum sulfide, we need to consider the stoichiometry of the reaction between aluminum sulfide and water to form aluminum hydroxide. Given the balanced chemical equation, we can calculate the molar mass of aluminum hydroxide and use it to convert the mass of aluminum sulfide to grams of aluminum hydroxide formed.
In the acid-base reaction where sodium hydroxide and sulfuric acid react, the formula is: H2SO4 + 2NaOH --> Na2SO4 + 2H2O. The coefficients shown are necessary to uphold the law of conservation of mass. So, if you have 17 moles of sulfuric acid, you will need twice as many moles of sodium hydroxide, so the answer is 34 moles NaOH.
In the reaction 4 moles of aluminum will react with 3 moles of oxygen to form 2 moles of aluminum oxide. Since we have 2.0 moles of aluminum, we would need (2.0 mol Al) x (3 mol O2 / 4 mol Al) = 1.5 moles of O2 to react with it.
The balanced chemical equation for the reaction between aluminum and chlorine is 2Al + 3Cl2 -> 2AlCl3. This means that for every 2 moles of aluminum that react, 2 moles of aluminum chloride are produced. Therefore, if 0.440 mol of aluminum is used, it will produce 0.440 mol of aluminum chloride.
To convert from molecules to moles, divide the number of molecules by Avogadro's number (6.022 x 10^23 molecules/mol). Therefore, 98.3 molecules of aluminum hydroxide would be approximately 0.163 moles.
To determine the grams of aluminum hydroxide produced from 14.2g of aluminum sulfide, first calculate the molar mass of aluminum sulfide by adding the atomic masses of aluminum and sulfur. Then, use the stoichiometry of the balanced chemical equation to find the molar ratio between aluminum sulfide and aluminum hydroxide. Next, convert the 14.2g of aluminum sulfide to moles, and then use the molar ratio to find the moles of aluminum hydroxide produced. Finally, convert the moles of aluminum hydroxide to grams using its molar mass.
1,99 grams of aluminum is equal to 0,0737 moles.
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------
There are 6 moles of sulfur present in 3 moles of aluminum sulfate, because aluminum sulfate has a 2:3 ratio of aluminum to sulfur.
To find the number of moles in 98.3 grams of aluminum hydroxide (Al(OH)₃), first calculate its molar mass. The molar mass of Al(OH)₃ is approximately 78.00 g/mol (26.98 g/mol for Al + 3 × 16.00 g/mol for O + 3 × 1.01 g/mol for H). Using the formula ( \text{moles} = \frac{\text{mass}}{\text{molar mass}} ), we have ( \text{moles} = \frac{98.3 , \text{g}}{78.00 , \text{g/mol}} \approx 1.26 , \text{moles} ). So, there are approximately 1.26 moles of aluminum hydroxide in 98.3 grams.
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
1,46 moles of aluminum fluoride contain 35,16848.10e23 atoms.
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
To determine the grams of aluminum hydroxide obtained from 17.2 grams of aluminum sulfide, we need to consider the stoichiometry of the reaction between aluminum sulfide and water to form aluminum hydroxide. Given the balanced chemical equation, we can calculate the molar mass of aluminum hydroxide and use it to convert the mass of aluminum sulfide to grams of aluminum hydroxide formed.
To determine how many moles of aluminum are produced from 33 grams, divide the given mass by the molar mass of aluminum, which is approximately 26.98 g/mol. So, 33 g / 26.98 g/mol ≈ 1.22 moles of aluminum are produced.
When aluminum oxide decomposes, it produces 2 moles of aluminum and 3 moles of oxygen for every mole of aluminum oxide. Therefore, for 26.5 moles of aluminum oxide decomposed, 3 * 26.5 = 79.5 moles of oxygen are produced.