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3 years ago

How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20

Chemistry

1 answer:

USPshnik [31]3 years ago

8 0

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and the last O atoms

Reaction

Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O

1. gives a coefficient

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O

2. make an equation

Zn : left = a, right =1 ⇒a=1

H : left = b, right = 2d⇒ b=2d (eq 1)

N : left = b, right = 2+c⇒b=2+c (eq 2)

O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)

From eq 1 and eq 3

3(2d)=6+c+d

6d=6+c+d

5d=6+c (eq 4)

From eq 2 and eq 3

3(2+c)=6+c+d

6+3c=6+c+d

2c=d (eq 5)

From eq 4 and eq 5

5(2c)=6+c

10c=6+c

9c=6

c = 2/3

input eq 5

d = 2 x 2/3

d = 4/3

input eq 1

b = 2 x 4/3

b = 8/3

The equation

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to

Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

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Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

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3 years ago

How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20​

How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20