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Mandarinka [93]
3 years ago
10

If 1.00 g of an unknown molecular compound contains 8.35  1021 molecules, what is its molar mass?

Chemistry
1 answer:
scoray [572]3 years ago
3 0
Answer is: molar mass of molecular compound is 71,95 g/mol.
N(compound) = 8,35·10²¹.
n(compound) = N(compound) ÷ Na.
n(compound) = 8,35·10²¹ ÷ 6,022·10²³ 1/mol.
n(compound) = 0,0139 mol.
m(compound) = 1,00 g.
M(compound) = m(compound) ÷ n(compound).
M(compound) = 1 g ÷ 0,0139 mol.
M(compound) = 71,95 g/mol.
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Answer:

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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:

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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

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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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