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3 years ago

C6H12O6(s) + 6O2(g) --> 6H2O(g) + 6CO2(g)

1 answer:

8 0

One
Balance the Equation.
This has been done for you or it is given. Anyway this step is finished, but it must always be done.

Two
Find the molar mass of C6H12O6
6C = 6 * 12 = 72
12H = 12*1 = 12
CO = 6 * 16 = 96
Mol Mass = 180 grams / mol

Three
Find the mols of C6H12O6
n = ???
Molar Mass = 180 grams / mol
given mass = 13.2 grams.

n = given mass / molar mass
n = 13.2 / 180
n = 0.07333333 mols.

Four
Find the mols CO2
1 mol C6H12O6 will produce 6 mols CO2
0.0733333 mols will produce x

1/0.073333 = 6/x Cross multiply
x = 0.073333 * 6
x = 0.4399 moles.

Five
Find the volume given the conditions for temperature and pressure are STP conditions.
PV = nRT
R = 0.082057
n = 0.43999
P = 1 atmosphere
T = 0 + 273 = 273
V = ???
1 * V = 0.43999 * 082057 * 273
V = 10.2 L
Answer: B

Note if you give out Brainly awards I'd sure appreciate one. This was a lot of typing

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An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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A molecule of C₂H₅OH has C-C, C-H, C-O, and O-H bonds.

A bond between A and B will be ionic if the difference between their electronegativities (ΔEN) is greater than 1.6.

$\begin{array}{ccc}\textbf{Bond} & \textbf{$\Delta$EN} & \textbf{Polarity}\\\text{C-C} & 2.55 - 2.55 = 0.00 & \text{Nonpolar covalent}\\\text{C-H} & 2.55 - 2.20 = 0.35 & \text{Nonpolar covalent}\\\text{C-O} & 3.44 - 2.55 = 0.89 & \text{Polar covalent}\\\text{O-H} & 3.44 - 2.20 = 1.24 & \text{Polar covalent}\\\end{array}$

No bond has a large enough ΔEN to be ionic.

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