Chemistry, 27.08.2020 21:01 oksanabkrot

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 from
the 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!

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Answer from: KBH3298

(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:

link

Answer from: bryan12376

(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.)

Answer from: Quest

c is the answer because when something is left after something evaporates that shows the chemical change.

Answer from: Quest

no, the red giant is the hottest with 2.00.000.000 degrees celsius. while the white dwarf only 25,000 degrees celsius.

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