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2 years ago

How many grams of CaO can be produced using 3.9 moles CaCO3?1 CaCO3 --> 1CaO + 1CO *

Chemistry

1 answer:

Lina20 [59]2 years ago

4 0

Answer: 218.4 g of $CaO$ will be produced from 3.9 moles of $CaCO_3$

Explanation:

The balanced chemical equation is:

$CaCO_3\rightarrow CaO+CO_2$

According to stoichiometry :

1 mole of $CaCO_3$ produce = 1 mole of $CaO$

Thus 3.9 moles of $CaCO_3$ will produce= $\frac{1}{1}\times 3.9=3.9moles$ of $CaO$

Mass of $CaO=moles\times {\text {Molar mass}}=3.9moles\times 56g/mol=218.4g$

Thus 218.4 g of $CaO$ will be produced from 3.9 moles of $CaCO_3$

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Metallic properties tend to increase in which direction on the periodic table

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Metallic properties head to the left.

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2 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago

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omeli [17]

From the balanced redox equation of the reaction, the coefficient of OH⁻ is 8.

<h3>What is the balanced redox equation of the reaction?</h3>

A redox equation is the equation of a redox reaction in which oxidation and reduction occurs simultaneously.

The given redox reaction takes place in a basic solution

The balanced redox equation of the reaction is given below:

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In conclusion, a redox equation is balanced when oxidation and reduction occur to the same extent.

Learn more about redox equations at: brainly.com/question/26750732

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1 year ago

A chemical company makes two brands of antifreeze. the first brand is 70% pure antifreeze, and the second brand is 95% pure anti

madreJ [45]

Answer is: 56 gallons of 70% antifreeze and 84 gallons of 95% antifreeze.

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