Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:
a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:
Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:
c) The leftover is computed as follows:
NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
Answer:
12.96 seconds
Explanation:
Assuming the reaction follows a first order
Rate = K[NOBr] = change in concentration of NOBr/time
K = 0.556 L mol^-1 s^-1
Change in concentration of NOBr = 0.32M - 0.039M = 0.281M
0.281/t = 0.556×0.039
t = 0.281/0.021684 = 12.96 seconds
Answer:
92.9%
Explanation:
You have been given the actual yield of the reaction. First, you need to find the theoretical yield of the reaction. To do this, you need to (1) convert grams Fe₂O₃ to moles Fe₂O₃ (via molar mass from periodic table values), then (2) convert moles Fe₂O₃ to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass).
Once you have found the theoretical yield, you need to use the percent yield equation to calculate the final answer. This number should have 3 sig figs to match the given values.
<u>(Step 1)</u>
Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)
Molar Mass (Fe₂O₃): 159.684 g/mol
1 Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)
Molar Mass (Fe): 55.845 g/mol
50.0 g Fe₂O₃ 1 mole 2 moles Fe 55.845 g
-------------------- x ------------------ x --------------------- x ---------------- = 35.0 g Fe
159.684 g 1 mole Fe₂O₃ 1 mole
<u>(Step 2)</u>
Actual Yield
Percent Yield = --------------------------- x 100%
Theoretical Yield
32.5 g Fe
Percent Yield = ---------------------- x 100% = 92.9%
35.0 g Fe
Done with an experiement and if he has enough data he can compare it and make his results more accurate please make me brainliest
This is only true for the titration of a strong acid with a strong base or vice versa. At the equivalence point of the titration of a weak base with a strong acid the pH is less than 7.00 and at the equivalence point of the titration of a weak acid with a strong base the pH is greater than 7.00.
