0
What else can I help you with?
How many grams of aluminum hydroxide are obtained from 17.2 grams of aluminum sulfide?
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.
How many atoms of aluminum are there in a sample of aluminum sulfide with a mass of 100 g?
Aluminum sulfide has a molar mass of 150.16 grams per mole. This means there are 0.666 moles present, or 4.01 E23 molecules. Each molecule of Al2S3 has 2 aluminum atoms, so there are 8.02 E23 atoms of aluminum present.
How many grams of sodium sulfide are formed if 1.90 of hydrogen sulfide is bubbled into a solution containing 2.53 of sodium hydroxide assuming that the sodium sulfide is made in 91.0 yield?
The equation for the reaction specified is 2 NaOH + H2S -> Na2S + H2O. Therefore, if the yield were 100 %, two formula masses of sodium hydroxide are required to produce one formula mass of sodium sulfide. The gram formula mass of NaOH is 40.00 and that of sodium sulfide is 78.04. The specified number of grams of sodium hydroxide corresponds to 2.53/40.00 or 0.06325 formula masses and therefore would provide half this many formula masses of sodium sulfide, for a mass of (0.06325)(78.04)/2.000 or 2.568 grams of sodium sulfide. Since the yield is specified as 91.0 %, the actual amount of sodium sulfide produced is 2.25 grams, to the justified number of significant digits.
If 328 grams of aluminum hydroxide react with excess hydrochloric acid how many grams of aluminum chloride can be formed?
Balanced equation first. ( you have been told the limiting reactant. ) Al(OH)3 + 3HCl >> AlCl3 + 3H2O Molar mass Al(OH)3 = 78.004 grams/// Molar mass AlCl3 = 133.33 grams Conversion. 328g Al(OH)3 (1mol Al(OH)3/78.004g)(1 mol AlCl3/1mol Al(OH)3)(133.33g AlCl3/1mol AlCl3) =560.64 grams produced
How many grams of iron II sulfide have to react with hydrochloric acid to generate 12 g of hydrogen sulfide?
To calculate the grams of iron II sulfide needed, we start by finding the moles of hydrogen sulfide produced. This is done by dividing the given mass of hydrogen sulfide by its molar mass. Then, we use the balanced chemical equation to determine that for every 4 moles of hydrogen sulfide, 1 mole of iron II sulfide is needed. From this, we find the grams of iron II sulfide required by multiplying the moles of iron II sulfide by its molar mass.
How many grams of aluminum hydroxide are obtained from 17.2 grams of aluminum sulfide?
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.
How many grams of aluminium hydroxide are obtained from 14.2g of aluminium sulfide?
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.
In making aluminum sulfide, 54 g of aluminum and 100 grams of sulfur make 150.3 g of aluminum sulfide and leftover sulfur. How much aluminum and sulfur should be used to make 150 g of aluminum sulfide with no leftovers?
Al2S3
How many grams of aluminum chloride are produced when 54 grams of aluminum react with chlorine gas?
266,86 g aluminium chloride are obtained.
How many atoms of aluminum are there in a sample of aluminum sulfide with a mass of 100 g?
Aluminum sulfide has a molar mass of 150.16 grams per mole. This means there are 0.666 moles present, or 4.01 E23 molecules. Each molecule of Al2S3 has 2 aluminum atoms, so there are 8.02 E23 atoms of aluminum present.
How many grams of sodium sulfide are formed if 1.90 of hydrogen sulfide is bubbled into a solution containing 2.53 of sodium hydroxide assuming that the sodium sulfide is made in 91.0 yield?
The equation for the reaction specified is 2 NaOH + H2S -> Na2S + H2O. Therefore, if the yield were 100 %, two formula masses of sodium hydroxide are required to produce one formula mass of sodium sulfide. The gram formula mass of NaOH is 40.00 and that of sodium sulfide is 78.04. The specified number of grams of sodium hydroxide corresponds to 2.53/40.00 or 0.06325 formula masses and therefore would provide half this many formula masses of sodium sulfide, for a mass of (0.06325)(78.04)/2.000 or 2.568 grams of sodium sulfide. Since the yield is specified as 91.0 %, the actual amount of sodium sulfide produced is 2.25 grams, to the justified number of significant digits.
If 328 grams of aluminum hydroxide react with excess hydrochloric acid how many grams of aluminum chloride can be formed?
Balanced equation first. ( you have been told the limiting reactant. ) Al(OH)3 + 3HCl >> AlCl3 + 3H2O Molar mass Al(OH)3 = 78.004 grams/// Molar mass AlCl3 = 133.33 grams Conversion. 328g Al(OH)3 (1mol Al(OH)3/78.004g)(1 mol AlCl3/1mol Al(OH)3)(133.33g AlCl3/1mol AlCl3) =560.64 grams produced
How many moles of aluminum oxide are present in a sample having a mass of 178.43g?
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
How much is 2.5 grams of aluminum?
How much is 5 grams of aluminum
Metallic aluminum can be obtained by electrolysis of aluminum fluoride dissolved in molten k3alf6. what mass of aluminum is obtained at the cathode if a 5 ampere current is applied for 4 hours?
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?
How many moles of aluminum are in 1.99 grams of aluminum?
1,99 grams of aluminum is equal to 0,0737 moles.
How do you Convert 42 grams AlOH3 to moles AlOH3?
That is aluminum hydroxide and as a polyatomic ion hydroxide needs to be in parenthesis. Thus; Al(OH)3 This shows aluminum's 3+ oxidation state and the three matching hydroxides 1- state. 42 grams Al(OH)3 ( 1 mole Al(OH)3/78.004 grams) = 0.538 moles Al(OH)3 As significant figures require; 0.54 moles Al(OH)3
