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Temka [501]
3 years ago
10

Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56 g of NaCl in 164.2 g of H2O. Express your answer to

four significant figures. View Available Hint(s)
Chemistry
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

"23.896%" is the right answer.

Explanation:

The given values are:

Mass of NaCl,

= 51.56 g

Mass of H₂O,

= 165.6 g

As we know,

⇒  Mass of solution = Mass \ of \ (NaCl+H_2O)

                                 = 51.56+164.2

                                 = 215.76 \ g

hence,

⇒ Mass \ percent =\frac{Mass \ of \ NaCl}{Mass \ of \ solution}\times 100

                           =\frac{51.56}{215.76}\times 100

                           =23.896 \ percent

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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
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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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