29. using the following equation, 2 naclo₃ → 2 nacl + 3 o₂, how many particles of naclo₃ are needed to…

29. using the following equation, 2 naclo₃ → 2 nacl + 3 o₂, how many particles of naclo₃ are needed to produce 68.3 l of oxygen?\n30. using the following equation, 2 kclo₃ → 2 kcl + 3 o₂, how many formula units of kclo₃ are decomposed if 6.95 g of kcl are produced?\n31. using the unbalanced equation, nh₃(g) + o₂(g) → no(g) + h₂o(l), how many molecules of ammonia are consumed if 28.14 g of water are produced?\n32. if iron (iii) oxide reacts with hydrogen gas to form iron and water, how many grams of water will be produced when 0.0155 moles of hydrogen gas react completely with iron (iii) oxide?\n33. using the following equation, 2 kclo₃ → 2 kcl + 3 o₂, how many grams of kcl are produced if 16.25 l of oxygen are produced?\n34. using the following equation, ch₄ + 2 o₂ → co₂ + 2 h₂o, how many moles of carbon dioxide are produced from the combustion of 110 g of methane?\n35. using the following equation, n₂ + 3 h₂ → 2 nh₃, what volume of nh₃ at stp is produced if 25 g of n₂ reacts with excess h₂?\n36. using the following equation, c₃h₈(g) + 5 o₂(g) → 3 co₂(g) + 4 h₂o(g), how many grams of co₂ are produced by a reaction that also produces 2.87 moles of water?

29. using the following equation, 2 naclo₃ → 2 nacl + 3 o₂, how many particles of naclo₃ are needed to produce 68.3 l of oxygen?\n30. using the following equation, 2 kclo₃ → 2 kcl + 3 o₂, how many formula units of kclo₃ are decomposed if 6.95 g of kcl are produced?\n31. using the unbalanced equation, nh₃(g) + o₂(g) → no(g) + h₂o(l), how many molecules of ammonia are consumed if 28.14 g of water are produced?\n32. if iron (iii) oxide reacts with hydrogen gas to form iron and water, how many grams of water will be produced when 0.0155 moles of hydrogen gas react completely with iron (iii) oxide?\n33. using the following equation, 2 kclo₃ → 2 kcl + 3 o₂, how many grams of kcl are produced if 16.25 l of oxygen are produced?\n34. using the following equation, ch₄ + 2 o₂ → co₂ + 2 h₂o, how many moles of carbon dioxide are produced from the combustion of 110 g of methane?\n35. using the following equation, n₂ + 3 h₂ → 2 nh₃, what volume of nh₃ at stp is produced if 25 g of n₂ reacts with excess h₂?\n36. using the following equation, c₃h₈(g) + 5 o₂(g) → 3 co₂(g) + 4 h₂o(g), how many grams of co₂ are produced by a reaction that also produces 2.87 moles of water?

Answer

29.

Explanation:

Step1: Find moles of O₂

At standard - temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. So, the number of moles of O₂, $n_{O_2}=\frac{V}{V_m}$, where $V = 68.3$ L and $V_m=22.4$ L/mol. $n_{O_2}=\frac{68.3}{22.4}\text{ mol}\approx3.05$ mol

Step2: Use mole - ratio from the balanced equation

From the equation $2NaClO_3\rightarrow2NaCl + 3O_2$, the mole - ratio of $NaClO_3$ to $O_2$ is $\frac{2}{3}$. Let the number of moles of $NaClO_3$ be $n_{NaClO_3}$. Then $n_{NaClO_3}=\frac{2}{3}n_{O_2}$. $n_{NaClO_3}=\frac{2}{3}\times3.05\text{ mol}\approx2.03$ mol

Step3: Convert moles to particles

Using Avogadro's number $N_A = 6.022\times10^{23}$ particles/mol, the number of particles of $NaClO_3$, $N=n_{NaClO_3}\times N_A$. $N = 2.03\times6.022\times10^{23}\text{ particles}\approx1.22\times10^{24}$ particles

Answer:

$1.22\times10^{24}$ particles

30.

Explanation:

Step1: Calculate moles of KCl

The molar mass of KCl, $M_{KCl}=39.1 + 35.45=74.55$ g/mol. The number of moles of KCl, $n_{KCl}=\frac{m}{M}$, where $m = 6.95$ g. $n_{KCl}=\frac{6.95}{74.55}\text{ mol}\approx0.0932$ mol

Step2: Use mole - ratio from the balanced equation

From the equation $2KClO_3\rightarrow2KCl + 3O_2$, the mole - ratio of $KClO_3$ to $KCl$ is 1:1. So, the number of moles of $KClO_3$ decomposed, $n_{KClO_3}=n_{KCl}=0.0932$ mol.

Step3: Convert moles to formula units

Using Avogadro's number $N_A = 6.022\times10^{23}$ formula units/mol, the number of formula units of $KClO_3$, $N=n_{KClO_3}\times N_A$. $N = 0.0932\times6.022\times10^{23}\text{ formula units}\approx5.61\times10^{22}$ formula units

Answer:

$5.61\times10^{22}$ formula units

31.

Explanation:

Step1: Balance the equation

The balanced equation is $4NH_3(g)+5O_2(g)\rightarrow4NO(g)+6H_2O(l)$.

Step2: Calculate moles of water

The molar mass of $H_2O$, $M_{H_2O}=2\times1.01 + 16.00 = 18.02$ g/mol. The number of moles of $H_2O$, $n_{H_2O}=\frac{m}{M}$, where $m = 28.14$ g. $n_{H_2O}=\frac{28.14}{18.02}\text{ mol}\approx1.56$ mol

Step3: Use mole - ratio from the balanced equation

The mole - ratio of $NH_3$ to $H_2O$ is $\frac{4}{6}=\frac{2}{3}$. Let the number of moles of $NH_3$ be $n_{NH_3}$. Then $n_{NH_3}=\frac{2}{3}n_{H_2O}$. $n_{NH_3}=\frac{2}{3}\times1.56\text{ mol}=1.04$ mol

Step4: Convert moles to molecules

Using Avogadro's number $N_A = 6.022\times10^{23}$ molecules/mol, the number of molecules of $NH_3$, $N=n_{NH_3}\times N_A$. $N = 1.04\times6.022\times10^{23}\text{ molecules}\approx6.26\times10^{23}$ molecules

Answer:

$6.26\times10^{23}$ molecules

32.

Explanation:

Step1: Write the balanced equation

The balanced equation is $Fe_2O_3+3H_2\rightarrow2Fe + 3H_2O$.

Step2: Use mole - ratio from the balanced equation

The mole - ratio of $H_2$ to $H_2O$ is 1:1. Given $n_{H_2}=0.0155$ mol, so $n_{H_2O}=n_{H_2}=0.0155$ mol.

Step3: Calculate mass of water

The molar mass of $H_2O$, $M_{H_2O}=18.02$ g/mol. The mass of water, $m = n\times M$. $m=0.0155\times18.02\text{ g}\approx0.279$ g

Answer:

$0.279$ g

33.

Explanation:

Step1: Find moles of O₂

At STP, $n_{O_2}=\frac{V}{V_m}$, where $V = 16.25$ L and $V_m = 22.4$ L/mol. $n_{O_2}=\frac{16.25}{22.4}\text{ mol}\approx0.725$ mol

Step2: Use mole - ratio from the balanced equation

From the equation $2KClO_3\rightarrow2KCl + 3O_2$, the mole - ratio of $KCl$ to $O_2$ is $\frac{2}{3}$. Let the number of moles of $KCl$ be $n_{KCl}$. Then $n_{KCl}=\frac{2}{3}n_{O_2}$. $n_{KCl}=\frac{2}{3}\times0.725\text{ mol}\approx0.483$ mol

Step3: Calculate mass of KCl

The molar mass of $KCl$, $M_{KCl}=74.55$ g/mol. The mass of $KCl$, $m=n\times M$. $m = 0.483\times74.55\text{ g}\approx36.0$ g

Answer:

$36.0$ g

34.

Explanation:

Step1: Calculate moles of methane

The molar mass of $CH_4$, $M_{CH_4}=12.01+4\times1.01 = 16.05$ g/mol. The number of moles of $CH_4$, $n_{CH_4}=\frac{m}{M}$, where $m = 110$ g. $n_{CH_4}=\frac{110}{16.05}\text{ mol}\approx6.85$ mol

Step2: Use mole - ratio from the balanced equation

From the equation $CH_4+2O_2\rightarrow CO_2 + 2H_2O$, the mole - ratio of $CH_4$ to $CO_2$ is 1:1. So, the number of moles of $CO_2$, $n_{CO_2}=n_{CH_4}=6.85$ mol

Answer:

$6.85$ mol

35.

Explanation:

Step1: Calculate moles of N₂

The molar mass of $N_2$, $M_{N_2}=2\times14.01 = 28.02$ g/mol. The number of moles of $N_2$, $n_{N_2}=\frac{m}{M}$, where $m = 25$ g. $n_{N_2}=\frac{25}{28.02}\text{ mol}\approx0.892$ mol

Step2: Use mole - ratio from the balanced equation

From the equation $N_2+3H_2\rightarrow2NH_3$, the mole - ratio of $N_2$ to $NH_3$ is 1:2. So, the number of moles of $NH_3$, $n_{NH_3}=2n_{N_2}$. $n_{NH_3}=2\times0.892\text{ mol}=1.784$ mol

Step3: Calculate volume of NH₃ at STP

At STP, $V = n\times V_m$, where $V_m = 22.4$ L/mol. $V=1.784\times22.4\text{ L}\approx40.0$ L

Answer:

$40.0$ L

36.

Explanation:

Step1: Use mole - ratio from the balanced equation

From the equation $C_3H_8(g)+5O_2(g)\rightarrow3CO_2(g)+4H_2O(g)$, the mole - ratio of $CO_2$ to $H_2O$ is $\frac{3}{4}$. Given $n_{H_2O}=2.87$ mol. Let the number of moles of $CO_2$ be $n_{CO_2}$. Then $n_{CO_2}=\frac{3}{4}n_{H_2O}$. $n_{CO_2}=\frac{3}{4}\times2.87\text{ mol}=2.1525$ mol

Step2: Calculate mass of CO₂

The molar mass of $CO_2$, $M_{CO_2}=12.01+2\times16.00 = 44.01$ g/mol. The mass of $CO_2$, $m=n\times M$. $m=2.1525\times44.01\text{ g}\approx94.7$ g

Answer:

$94.7$ g