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Examine the following balanced chemical equation.rnrnalcl3(aq) + 3agno3(aq) --> Al(NO3)3(aq) + 3agcl(s) rnrndetermine the mass in grams of AgCl(s) produced, if you start off with 0.500 g of AlCl3 and 0.750 g of AgNO3.
Accepted Answer
To determine the mass of AgCl(s) produced, we first need to find the limiting reactant. To do this, we can compare the moles of each reactant to the stoichiometry of the balanced chemical equation:
AlCl3(aq) + 3AgNO3(aq) -→ Al(NO3)3(aq) + 3AgCl(s)
Start with 0.500 g AlCl3:
moles AlCl3 = 0.500 g / 133.34 g/mol = 0.00374 mol
Start with 0.750 g AgNO3:
moles AgNO3 = 0.750 g / 169.87 g/mol = 0.00442 mol
Comparing the moles of each reactant to the stoichiometry, we see that AlCl3 is the limiting reactant because it produces the least moles of product.
Now that we know the limiting reactant, we can use the stoichiometry of the balanced chemical equation to calculate the moles of AgCl(s) produced:
moles AgCl = 3 * moles AlCl3 = 3 * 0.00374 mol = 0.0112 mol
Finally, we can convert the moles of AgCl to grams:
mass AgCl = moles AgCl * molar mass AgCl
mass AgCl = 0.0112 mol * 143.32 g/mol = 1.60 g
Therefore, the mass of AgCl(s) produced is 1.60 g.
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