Asked by amanda
Aluminum nitrate and ammonium chloride react to form alluminum chloride, nitrogen, and water. what mass of aluminum chloride is present after 43.0 g of aluminum nitrate and 43.0 g ammonium chloride have reacted completely?
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Answered by
DrBob222
The is a limiting reagent problem. We know that because BOTH reactants are given in the problem.
1. Write and balance the equation.
2a. Convert 43.0 g Al(NO3)3 to moles. moles = grams/molar mass.
2b. Convert 43.0 g NH4Cl to moles the same way.
3a. Using the coefficients in the balanced equation, convert moles Al(NO3)3 to moles AlCl3.
3b. Same procedure, convert moles NH4Cl to moles AlCl3.
3c. The moles from 3a and 3b likely will not be the same and one of them must be wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
4.Using the smaller value, convert moles of the product to grams. g = moles x molar mass.
1. Write and balance the equation.
2a. Convert 43.0 g Al(NO3)3 to moles. moles = grams/molar mass.
2b. Convert 43.0 g NH4Cl to moles the same way.
3a. Using the coefficients in the balanced equation, convert moles Al(NO3)3 to moles AlCl3.
3b. Same procedure, convert moles NH4Cl to moles AlCl3.
3c. The moles from 3a and 3b likely will not be the same and one of them must be wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
4.Using the smaller value, convert moles of the product to grams. g = moles x molar mass.
Answered by
mart
3.02
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