Answer:
endothermic
Explanation:
condensate is an endothermic change
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Answer:
144 g
Explanation:
Use the mole ratio of 4 mol CO2 for every 9 mol O2 to convert from mol O2 to mol CO2. Then use the molar mass of CO2 to convert from mol of CO2 to grams of CO2.
7.34 mol O2 • (4 mol CO2 / 9 mol O2) • (44.01 g CO2 / 1 mol CO2) = 144 g CO2
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