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Arada [10]
3 years ago
6

A solution is made by mixing 15.0 g of sr(oh)2 and 55.0 ml of 0.200 m hno3.

Chemistry
1 answer:
noname [10]3 years ago
4 0

 Letter A: The balanced equation is:

Sr(OH)2 (s) + 2 HNO3 (aq) → Sr(NO3)2 (aq) + 2 H2O 


Letter B: The solution is:


(15.0 g Sr(OH)2) / (121.6358 g Sr(OH)2/mol) = 0.12332 mol Sr(OH)2 
(0.0550 L) x (0.200 mol/L HNO3) = 0.011 mol HNO3 


0.011 mole of HNO3 would respond entirely with 0.011 x (1/2) = 0.0055 mole of Sr(OH)2 but there is more Sr(OH)2 existing than that, so Sr(OH)2 is in excess and HNO3 is the limiting reactant. 


( 0.011 mol HNO3) x (1/2) = 0.0055 mol Sr(NO3)2 


Since the NO3{-} ions continued in solution during, the concentration of the NO3{-} ions didn't change during the reaction, so they are still 0.200 M. 
The concentration of Sr{+2} ions is (0.0055 mol Sr(NO3)2) / (55.0 mL) = 0.100 M 
Since HNO3 is the limiting reactant, no H{+} ions continue at the end of the reaction. 

Letter C: Since all of the acid reacted and there was still Sr(OH)2 left over, the resulting solution would be basic.

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3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

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:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

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:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

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:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

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.

7 0
3 years ago
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