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Aluminum nitrite & ammonium chloride react to form aluminum chloride, nitrogen gas and water. What amount of aluminum chloride is present after 43g of ammonium chloride have reacted completely?
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1. Write the balanced chemical equation.
2. Convert 43 g NH4Cl to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles NH4Cl to moles AlCl3.
4. Now convert moles AlCl3 to grams. g = moles x molar mass.
Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of each substance is present after 62.3 g aluminum nitrite and 51.2 g ammonium chloride react completely?
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To find the amount of aluminum chloride formed, we need to use the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction:
2Al(NO3)3 + 3NH4Cl -> 2AlCl3 + 3N2 + 12H2O
From the balanced chemical equation, we can see that for every 3 moles of ammonium chloride (NH4Cl) reacted, 2 moles of aluminum chloride (AlCl3) are formed.
Now, let's calculate the number of moles of ammonium chloride (NH4Cl) using the given mass:
molar mass of NH4Cl = 14 + 4 + 1 x 4 = 53 g/mol
moles of NH4Cl = mass / molar mass = 43g / 53 g/mol
Using the mole ratio from the balanced equation, we can now calculate the moles of aluminum chloride (AlCl3) formed:
moles of AlCl3 = moles of NH4Cl x (2 moles of AlCl3 / 3 moles of NH4Cl)
Finally, to find the amount of aluminum chloride in grams, we multiply the moles of AlCl3 by its molar mass:
mass of AlCl3 = moles of AlCl3 x molar mass of AlCl3
Make sure to plug in all the actual values into the calculations to obtain the numerical value of the amount of aluminum chloride formed.