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Mashcka [7]
3 years ago
15

Please Help!!!!!!!

Chemistry
1 answer:
hichkok12 [17]3 years ago
8 0

3,1,3,1

Explanation:

 Reaction equation:

    HBr    +    Al(OH)₃   ⇒    H₂O    +     AlBr₃

To balance an equation we make sure that the same number of atoms appears on both sides of the expression.

  HBr    +    Al(OH)₃   ⇒    H₂O    +     AlBr₃

Let us a simple mathematical method to solve this problem

  aHBr    +    bAl(OH)₃   ⇒    cH₂O    +     dAlBr₃

  a, b , c and d are the coefficients that will make the equation balanced;

 Conserving H;  a + 3b = 2c

                      Br;  a  = 3d

                      Al;   b  = d

                       O;    3b  = c

    let us make a = 1

                         d = \frac{1}{3}

                         b = \frac{1}{3}

                          c = 1

multiply through by 3;

     a = 3, b = 1, c= 3 and d = 1

learn more:

Balanced equation brainly.com/question/9325293

#learnwithBrainly

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Approximately 81.84\%.

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

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\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}.

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\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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