If I have one mole of sulfur, how many atoms would that be?

Chemistry

1 answer:

krok68 [10]3 years ago

7 0

Answer:

Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1

Explanation:

because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.

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3. How many grams of water will be produced from 15.0 grams of Methane? *<br> CH4 +2 02 → CO2 + 2H2O

jok3333 [9.3K]

CH4 + 2O2 = CO2 + 2H2O

According to molar weights :

16 gm CH4 + 64 gm O2 = 44 gm CO2 + 36 gm H2O

Since 16 gm CH4 produce 36 gm H2O

Hence 2.5 gmCH4 produce 36×2.5/16 gm H2O

= 5.265 gm of H2O
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Related Questions (More Answers Below)

8 0

2 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$