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3 years ago

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Wine goes bad soon after opening because the ethanol CH3CH2OH dissolved in it reacts with oxygen O2 gas to form water and aqueou

s acetic acid CH3COOH, the main ingredient in vinegar. Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

1 answer:

lara [203]3 years ago

6 0

Answer:

1.70 mol of ethanol is needed

Explanation:

First, we have to write the balanced equation for reaction of ethanol with oxygen to produce acetic acid and water

Balanced reaction:

$CH_{3}CH_{2}OH+O_{2}\rightarrow CH_{3}COOH+H_{2}O$

According to balanced equation, 1 mol of ethanol produces 1 mol of $H_{2}O$

So, moles of ethanol needed to produce 1.70 mol of water = 1.70 mol

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What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M<br> solution?

love history [14]

Answer:

Approximately $19.1\; \rm g$ .

Explanation:

<h3>Number of moles of formula units of magnesium sulfate required to make the solution</h3>

The unit of concentration in this question is " $\rm M$ ". That's equivalent to " $\rm mol \cdot L^{-1}$ " (moles per liter.) In other words:

$c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}$ .

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

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Calculate the number of moles of $\rm MgSO_4$ formula units required to make this solution:

$\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}$ .

<h3>Mass of magnesium sulfate in the solution</h3>

Look up the relative atomic mass data of $\rm Mg$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm Mg$ : $24.305$ .
$\rm S$ : $\rm 32.06$ .
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Calculate the formula mass of $\rm MgSO_4$ using these values:

$M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}$ .

Using this formula mass, calculate the mass of that (approximately) $0.159\; \rm mol$ of $\rm MgSO_4$ formula units:

$\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}$ .

Therefore, the mass of $\rm MgSO_4$ required to make this solution would be approximately $19.1\; \rm g$ .

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3 years ago

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Ierofanga [76]

Answer:

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Explanation:

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