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UkoKoshka [18]
3 years ago
5

Calcium and Chlorine are reacted together. What is the product of this reaction?

Chemistry
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

Calcium chloride (CaCl2)

Explanation:

Ca + Cl2 > CaCl2

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Assuming the substance is one of those listed in the table presented in this lesson, what is the identity of the substance?
zepelin [54]

The identity of the substance is Iron.

<h3>Properties of Iron</h3><h3>Chemical properties</h3>
  • The element iron has the atomic number 26 and the symbol Fe
  • Electronic configuration of Fe is [AR] 3d6 4s2
  • The atomic weight of Iron is 55.847
  • The element iron belong to the group VIII of the periodic table
  • It is the fourth most prevalent element in the crust of the world
  • It is a highly reactive element as it gets rusted readily in the moisture of air
<h3>Physical properties</h3>
  • It is a heavy metal in the first transition series  
  • The color of iron is silvery grey
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Learn more about periodic table on

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8 0
2 years ago
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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
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.

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My answer to this question is C

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