Chemistry: A reaction between 7.0 g of copper(II) oxide and 50 mL of 0.20 M nitric acid produces copper(II) nitrate, Cu(NO3)2 and water.(c) Determine the limiting reactant.(d) Calculate the mass of excess reactant after the reaction.(ANS: 6.6068g)(e) Determine the percentage yield if the actual mass of copper (II) nitrate obtained fromthe reaction is 0.85 g.(ANS: 90.64%)How to get the mass of HNO3 from here? I only managed to get mass of NO3 based on the molarity formula. thanks!

A reaction between 7.0 g of copper(II) oxide and

Ответ:

anthonygeorge7

31.01.2022

Chemistry

(d) The mass of the excess reactant after the reaction is approximately 6.6 grams

(e) The percentage yield of copper (II) nitrate from the reaction is approximately 90.64%

The reason the above values are the correct values is as follows;

The given parameters are;

The mass of copper(II)oxide in the reaction = 7.0 g

The volume of the 0.20 M nitric acid, HNO₃ = 50 mL

The molar mass of CuO = 79.545 g/mol

Number of moles = Mass/(Molar mass)

The number of moles of CuO = (7 g)/79.545 g/mol ≈ 0.088 moles

50 mL of 0.20 M HNO₃, contains 50/1000 × 0.2 = 0.01 moles of HNO₃

The chemical equation for the reaction is CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O

Therefore;

One mole of CuO reacts with two moles of HNO₃ to produce one mole of Cu(NO₃)₂ and one mole of H₂O

Therefore, 0.088 moles of CuO reacts with 2 × 0.088 = 0.176 moles of HNO₃

Given that there is only 0.01 moles of HNO₃, the limiting reactant is the HNO₃, which is not enough to completely react with the CuO which is the excess reactant

(d) The mass of the CuO that reacts with the 0.01 moles of HNO₃ is given as follows;

1 mole of CuO reacts with 2 moles HNO₃

0.01 moles of HNO₃ will react with 0.01/2 = 0.005 moles of CuO

Mass = Number of moles × Molar mass

The mass of 0.005 moles of CuO = 0.005 moles × 79.545 g/mol = 0.397725 grams

The mass of the CuO left = Initial mass - Reacting mass

∴ The mass of the CuO left = 7 grams - 0.397725 grams = 6.602275 grams

The mass of the excess reactant (CuO) after the reaction ≈ 6.6 grams

(e) The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = Half the number of moles of HNO₃ in the reaction

The number of moles of HNO₃ in the reaction = 0.01 moles

∴ The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = (1/2) × 0.01 moles = 0.005 moles

The molar mass of Cu(NO₃)₂ = 187.56 g/mol

The theoretical mass of Cu(NO₃)₂ produced = 0.005 moles × 187.56 g/mol = 0.9378 grams

The actual yield of copper (II) nitrate is 0.84 g

$Percentage \ yield = \mathbf{\dfrac{Actual \ yield}{Theoretical \ yield} \times 100}$

Therefore;

$\% \ yield \ of \ Cu(NO_3)_2 = \dfrac{0.85 \, g}{0.9378 \, g} \times 100 \approx 90.64 \%$

The percentage yield of copper (II) nitrate, %yield ≈ 90.64%

Learn more about percentage yield from chemical reactions here:

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Ответ:

itsstaytay200

31.01.2022

Chemistry

(d) Approximately $6.6\; \rm g$ of $\rm CuO\, (s)$ will be in excess.

(e) The percentage yield of $\rm Cu(NO_3)_2$ is approximately $91\%$ . (Rounded to two significant figures, as in other quantities in the question.)

Explanation:

Start with the balanced chemical equation:

$\rm CuO\, (s) + 2\; HNO_3\, (aq) \to Cu(NO_3)_2\, (aq) + H_2O\, (l)$ .

Look up relevant relative atomic mass data on a modern periodic table:

$\rm Cu$ : 63.546

. $\rm O$ : $\rm 15.999$ . $\rm H$ : 1.008

. $\rm N$ : 14.007

Calculate the formula mass of the species:

$M(\mathrm{CuO}) = 63.546 + 15.999 = 79.545\; \rm g\cdot mol^{-1}$ . $M(\mathrm{Cu(NO_3)_2}) = 63.546 + 2\times (14.007 + 3 \times 15.999) = 187.554\; \rm g\cdot mol^{-1}$ .Limiting Reactant

There are two reactants in this reaction: $\rm CuO$ and $\rm HNO_3\, (aq)$ . Assume that $\rm CuO\!$ is the limiting one. In other words, assume that all the $\rm CuO\!\!$ is consumed before $\rm HNO_3\, (aq)\!$ was.

Consider: how many moles of $\rm HNO_3\, (aq)\!\!$ would be required to convert all that $7.0\; \rm g$ of $\rm CuO\!\!\!$ to $\rm Cu(NO_3)_2$ ?

Calculate the number of moles of $\rm CuO$ formula units in $7.0\;\rm g$ of

$\displaystyle n({\rm CuO}) = \frac{m}{M} = \frac{7.0\; \rm g}{79.545\; \rm g \cdot mol^{-1}} \approx 0.088001\; \rm mol$ .

Note the ratio between the coefficients of $\rm CuO\, (s)$ and $\rm HNO_3\, (aq)$ :

$\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{CuO\, (s)})} = \frac{2}{1} = 2$ .

Therefore:

$\begin{aligned}&n(\text{$\mathrm{HNO_3\, (aq)}$, consumed (under assumption)})\\ &= \frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\cdot n(\mathrm{CuO\, (s)}) \\ &\approx 2 \times 0.088001\; \rm mol \approx 0.17600\; \rm mol\end{aligned}$ .

On the other hand, how many moles of $\rm HNO_3\, (aq)$ are actually available?

Convert the volume of that $\rm HNO_3\, (aq)$ solution to the standard unit (liter.)

$V(\rm HNO_3\, (aq)) = 50\; \rm mL = 0.050\; \rm L$ .

Calculate the number of moles of $\rm HNO_3\, (aq)$ in that $0.20\; \rm M$ solution:

$n({\rm HNO_3\, (aq)}) = c\cdot V = 0.050\; \rm L \times 0.20\; \rm mol \cdot L^{-1} = 0.010\; \rm mol$ .

Apparently, the quantity of $\rm HNO_3\, (aq)$ required exceeded the quantity that is available. Therefore, the assumption is invalid, and $\rm CuO$ cannot be the limiting reactant. At the same time,

Mass of the reactant in excess

Since it is now known that all that $0.010\; \rm mol$ of $\rm HNO_3\, (aq)$ will be consumed, apply that coefficient ratio again to obtain the quantity of $\rm CuO$ consumed in this reaction:

$\begin{aligned}&n(\text{$\mathrm{CuO}$, consumed})\\ &= n(\text{$\mathrm{HNO_3\, (aq)}$, consumed}) \left/\frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\right. \\ &\approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol\end{aligned}$ .

It was already shown that the formula mass of $\rm CuO$ is (approximately) $79.545\; \rm g \cdot mol^{-1}$ . Therefore, the mass of that $0.0050\; \rm mol$ formula units of $\rm CuO\!$ would be:

$\begin{aligned}& m(\text{$\rm CuO$, consumed}) \\&= n \cdot M \approx 0.0050\; \rm mol \times 79.545\; \rm g\cdot mol^{-1} \\&\approx 0.39773\; \rm g\end{aligned}$ .

Before the reaction, $7.0\; \rm g$ of $\rm CuO$ is available. Therefore, all that $7.0\; \rm g - 0.39773\; \rm g \approx 6.6\; \rm g$ of $\rm CuO\!$ would be in excess.

Percentage Yield

Similarly:

$\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{Cu(NO_3)_2\, (aq}))} = \frac{2}{1} = 2$ .

Apply this ratio to find the theoretical yield of $\rm Cu(NO_3)_2$ :

$n(\text{$\mathrm{Cu(NO_3)_2\, (aq)}$, theoretical yield}) \approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol$ .

Find the mass of that $0.0050\; \rm mol$ of $\rm Cu(NO_3)_2$ formula units using its formula mass:

$m(\text{$\rm Cu(NO_3)_2\, (aq)$, theoretical yield}) \approx 0.93777\; \rm g$ .

Calculate the percentage yield given that the actual yield is $0.85\; \rm g$ :

$\begin{aligned}&\text{Percentage Yield} \\ &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\% \\ &= \frac{0.85\; \rm g}{0.93777\; \rm g} \times 100\% \approx 0.91\% \end{aligned}$ .

(Rounded to two significant figures.)

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Ответ:

josiesolomonn1615

06.02.2023

Chemistry

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Explanation:

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