Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. if 10.6 grams of aluminum hydroxide is produced, how many grams of water were reacted?

There reacts 7.35 grams of H2O

Explanation:

Step 1: Data given

Mass of aluminium hydroxide = 10.6 grams

Step 2: The balanced equation

Al2S3(s) + 6H2O(l) → 2Al(OH)3(s) + 3H2S(g)

Step 3: Calculate number of moles Al(OH)3

Moles Al(OH)3 = mass Al(OH)3 / molar mass Al(OH)3

Moles Al(OH)3 = 10.6 grams / 78.00 g/mol

Moles Al(OH)3 = 0.136 moles

Step 4: Calculate moles of H2O

For 1 mol of Al2S3 we need 6 moles of H2O to produce 2 moles of Al(OH)3 and 3 moles of H2S

For 0.136 moles of Al(OH)3 we need 3*0.136 = 0.408 moles of H2O

Step 5: Calculate mass of H2O

Mass H2O = moles H2O * molar mass of H2O

Mass H2O = 0.408 * 18.02 = 7.35 grams

There reacts 7.35 grams of H2O

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explanation:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :