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1 year ago

100 points! please help me will mark brainliest

Chemistry

2 answers:

aalyn [17]1 year ago

7 0

S + O2 → SO2

Moles of S:

11/32.06 = 0.3431 mol

Moles of O2:

44/15.999 = 2.75 mol

Sulfur will be the limiting reagent.

Oxygen will be the excess reagent

2.75 - 0.3431 = 2.4071 mols of oxygen leftover

2.4071 * 15.999 = 38.511

38.511 grams of oxygen will be leftover

True [87]1 year ago

5 0

Reaction

S + O₂ ⇒ SO₂

mole S = 11 : 32 g/mol = 0.34375

mole O₂ = 44 : 32 g/mol = 1.375

S as a limiting reactant (smaller) and converted all to sulfur dioxide

mole O₂ (unreacted) = 1.375 - 0.34375 = 1.03125

mass O₂ (unreacted) = 1.03125 x 32 = 33 g

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The energy produced equals 140.760 kJ

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3 years ago

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Elena-2011 [213]

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Hope this is the answer you are looking for mate!

If it is correct then please mark my answer as the brailiest! :)

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3 years ago

The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo

noname [10]

Answer:

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

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P1 is the initial pressure

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Since we're finding the initial volume

$V_1 = \frac{P_2V_2}{P_1} \\$

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We have the final answer as

Hope this helps you

8 0

2 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.

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2 years ago

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