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3 years ago

8

Propane, c3h8, burns to produce carbon dioxide and water by the equation below. what is the coefficient in front of the carbon d

ioxide in the balanced equation? c3h8 + o2 \rightarrow→→ co2 + h2o

1 answer:

jeka57 [31]3 years ago

8 0

Answer:

3

Explanation:

The unbalanced combustion reaction of the oxidation of propane is shown below as:-

$C_3H_8+O_2\rightarrow CO_2+H_2O$

On the left hand side,

There are 3 carbon atoms and 8 hydrogen atoms and 2 oxygen atoms

On the right hand side,

There are 1 carbon atom and 2 hydrogen atoms and 3 oxygen atoms

Thus,

Right side, $CO_2$ must be multiplied by 3 and $H_2O$ by 4 so to balance carbon and hydrogen.

Left side, $O_2$ is multiplied by 5 so to balance the whole reaction.

Thus, the balanced reaction is:-

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

<u>Coefficient in front of the carbon dioxide in the balanced equation - 3</u>

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How many solutions are there in the following equation.

givi [52]

When they ask you for the solution they are usually asking for x. So solve for X.
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3 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

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Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

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2 years ago

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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