17. using the following equation, 2 naclo₃ → 2 nacl + 3 o₂, how many grams of naclo₃ must decompose to…

17. using the following equation, 2 naclo₃ → 2 nacl + 3 o₂, how many grams of naclo₃ must decompose to produce 114.25 g of nacl?\n18. using the unbalanced equation, nh₃(g) + o₂(g) → no(g) + h₂o(l), how many grams of water are produced from reacting 9.25 g of ammonia\n19. in the following equation, nh₄no₃(s) → n₂o(g) + 2 h₂o(l), if 50.6 g of nh₄no₃ decompose how much n₂o is produced?\n20. given the following unbalanced equation, h₂ + o₂ → h₂o, how many grams of hydrogen gas is needed to react with 21 g of oxygen gas?\n21. using the following equation, 3 agno₃ + alcl₃ → 3 agcl + al(no₃)₃, what is the mass of agcl that is produced from reacting 327 g of alcl₃?\n22. using the following unbalanced equation, ch₄ + o₂ → co₂ + h₂o, what is the mass of water produced from the reaction of 65 grams of methane?\n23. using the following unbalanced equation, kclo₃ → kcl + o₂, how many grams do kclo3 decompose if you produce 8.24 x 10²⁵ molecules of oxygen?\n24. in the following equation, nh₄no₃(s) → n₂o(g) + 2 h₂o(l), if 78.2 g of nh₄no₃ decompose how much water is produced?

17. using the following equation, 2 naclo₃ → 2 nacl + 3 o₂, how many grams of naclo₃ must decompose to produce 114.25 g of nacl?\n18. using the unbalanced equation, nh₃(g) + o₂(g) → no(g) + h₂o(l), how many grams of water are produced from reacting 9.25 g of ammonia\n19. in the following equation, nh₄no₃(s) → n₂o(g) + 2 h₂o(l), if 50.6 g of nh₄no₃ decompose how much n₂o is produced?\n20. given the following unbalanced equation, h₂ + o₂ → h₂o, how many grams of hydrogen gas is needed to react with 21 g of oxygen gas?\n21. using the following equation, 3 agno₃ + alcl₃ → 3 agcl + al(no₃)₃, what is the mass of agcl that is produced from reacting 327 g of alcl₃?\n22. using the following unbalanced equation, ch₄ + o₂ → co₂ + h₂o, what is the mass of water produced from the reaction of 65 grams of methane?\n23. using the following unbalanced equation, kclo₃ → kcl + o₂, how many grams do kclo3 decompose if you produce 8.24 x 10²⁵ molecules of oxygen?\n24. in the following equation, nh₄no₃(s) → n₂o(g) + 2 h₂o(l), if 78.2 g of nh₄no₃ decompose how much water is produced?

Answer

Question 17

Explanation:

Step1: Calculate molar - mass of NaCl

The molar mass of Na (sodium) is approximately $22.99\ g/mol$, and the molar mass of Cl (chlorine) is approximately $35.45\ g/mol$. So, the molar mass of NaCl, $M_{NaCl}=22.99 + 35.45=58.44\ g/mol$.

Step2: Calculate moles of NaCl

The number of moles of NaCl, $n_{NaCl}=\frac{m_{NaCl}}{M_{NaCl}}=\frac{114.25\ g}{58.44\ g/mol}\approx1.955\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $2NaClO_3\rightarrow2NaCl + 3O_2$, the mole - ratio of $NaClO_3$ to $NaCl$ is $1:1$. So, the number of moles of $NaClO_3$, $n_{NaClO_3}=n_{NaCl}=1.955\ mol$.

Step4: Calculate molar - mass of $NaClO_3$

The molar mass of Na is $22.99\ g/mol$, Cl is $35.45\ g/mol$, and O is $16.00\ g/mol$. So, $M_{NaClO_3}=22.99+35.45 + 3\times16.00=106.44\ g/mol$.

Step5: Calculate mass of $NaClO_3$

The mass of $NaClO_3$, $m_{NaClO_3}=n_{NaClO_3}\times M_{NaClO_3}=1.955\ mol\times106.44\ g/mol\approx208\ g$.

Answer:

$208\ g$

Question 18

First, balance the equation: $4NH_3(g)+5O_2(g)\rightarrow4NO(g)+6H_2O(l)$.

Explanation:

Step1: Calculate molar - mass of $NH_3$

The molar mass of N is $14.01\ g/mol$ and H is $1.01\ g/mol$. So, $M_{NH_3}=14.01 + 3\times1.01=17.04\ g/mol$.

Step2: Calculate moles of $NH_3$

The number of moles of $NH_3$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{9.25\ g}{17.04\ g/mol}\approx0.543\ mol$.

Step3: Determine mole - ratio from the balanced equation

The mole - ratio of $NH_3$ to $H_2O$ is $4:6 = 2:3$. So, the number of moles of $H_2O$, $n_{H_2O}=\frac{3}{2}n_{NH_3}=\frac{3}{2}\times0.543\ mol = 0.8145\ mol$.

Step4: Calculate molar - mass of $H_2O$

The molar mass of H is $1.01\ g/mol$ and O is $16.00\ g/mol$. So, $M_{H_2O}=2\times1.01+16.00 = 18.02\ g/mol$.

Step5: Calculate mass of $H_2O$

The mass of $H_2O$, $m_{H_2O}=n_{H_2O}\times M_{H_2O}=0.8145\ mol\times18.02\ g/mol\approx14.7\ g$.

Answer:

$14.7\ g$

Question 19

Explanation:

Step1: Calculate molar - mass of $NH_4NO_3$

The molar mass of N is $14.01\ g/mol$, H is $1.01\ g/mol$, and O is $16.00\ g/mol$. So, $M_{NH_4NO_3}=2\times14.01+4\times1.01 + 3\times16.00=80.04\ g/mol$.

Step2: Calculate moles of $NH_4NO_3$

The number of moles of $NH_4NO_3$, $n_{NH_4NO_3}=\frac{m_{NH_4NO_3}}{M_{NH_4NO_3}}=\frac{50.6\ g}{80.04\ g/mol}\approx0.632\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $NH_4NO_3(s)\rightarrow N_2O(g)+2H_2O(l)$, the mole - ratio of $NH_4NO_3$ to $N_2O$ is $1:1$. So, the number of moles of $N_2O$, $n_{N_2O}=n_{NH_4NO_3}=0.632\ mol$.

Step4: Calculate molar - mass of $N_2O$

The molar mass of N is $14.01\ g/mol$ and O is $16.00\ g/mol$. So, $M_{N_2O}=2\times14.01+16.00 = 44.02\ g/mol$.

Step5: Calculate mass of $N_2O$

The mass of $N_2O$, $m_{N_2O}=n_{N_2O}\times M_{N_2O}=0.632\ mol\times44.02\ g/mol\approx27.8\ g$.

Answer:

$27.8\ g$

Question 20

First, balance the equation: $2H_2+O_2\rightarrow2H_2O$.

Explanation:

Step1: Calculate molar - mass of $O_2$

The molar mass of O is $16.00\ g/mol$, so $M_{O_2}=2\times16.00 = 32.00\ g/mol$.

Step2: Calculate moles of $O_2$

The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{21\ g}{32.00\ g/mol}\approx0.656\ mol$.

Step3: Determine mole - ratio from the balanced equation

The mole - ratio of $H_2$ to $O_2$ is $2:1$. So, the number of moles of $H_2$, $n_{H_2}=2n_{O_2}=2\times0.656\ mol = 1.312\ mol$.

Step4: Calculate molar - mass of $H_2$

The molar mass of H is $1.01\ g/mol$, so $M_{H_2}=2\times1.01 = 2.02\ g/mol$.

Step5: Calculate mass of $H_2$

The mass of $H_2$, $m_{H_2}=n_{H_2}\times M_{H_2}=1.312\ mol\times2.02\ g/mol\approx2.65\ g$.

Answer:

$2.65\ g$

Question 21

Explanation:

Step1: Calculate molar - mass of $AlCl_3$

The molar mass of Al is $26.98\ g/mol$ and Cl is $35.45\ g/mol$. So, $M_{AlCl_3}=26.98+3\times35.45 = 133.33\ g/mol$.

Step2: Calculate moles of $AlCl_3$

The number of moles of $AlCl_3$, $n_{AlCl_3}=\frac{m_{AlCl_3}}{M_{AlCl_3}}=\frac{327\ g}{133.33\ g/mol}\approx2.45\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $3AgNO_3+AlCl_3\rightarrow3AgCl + Al(NO_3)3$, the mole - ratio of $AlCl_3$ to $AgCl$ is $1:3$. So, the number of moles of $AgCl$, $n{AgCl}=3n_{AlCl_3}=3\times2.45\ mol = 7.35\ mol$.

Step4: Calculate molar - mass of $AgCl$

The molar mass of Ag is $107.87\ g/mol$ and Cl is $35.45\ g/mol$. So, $M_{AgCl}=107.87+35.45 = 143.32\ g/mol$.

Step5: Calculate mass of $AgCl$

The mass of $AgCl$, $m_{AgCl}=n_{AgCl}\times M_{AgCl}=7.35\ mol\times143.32\ g/mol\approx1050\ g$.

Answer:

$1050\ g$

Question 22

First, balance the equation: $CH_4+2O_2\rightarrow CO_2 + 2H_2O$.

Explanation:

Step1: Calculate molar - mass of $CH_4$

The molar mass of C is $12.01\ g/mol$ and H is $1.01\ g/mol$. So, $M_{CH_4}=12.01+4\times1.01 = 16.05\ g/mol$.

Step2: Calculate moles of $CH_4$

The number of moles of $CH_4$, $n_{CH_4}=\frac{m_{CH_4}}{M_{CH_4}}=\frac{65\ g}{16.05\ g/mol}\approx4.05\ mol$.

Step3: Determine mole - ratio from the balanced equation

The mole - ratio of $CH_4$ to $H_2O$ is $1:2$. So, the number of moles of $H_2O$, $n_{H_2O}=2n_{CH_4}=2\times4.05\ mol = 8.1\ mol$.

Step4: Calculate molar - mass of $H_2O$

The molar mass of H is $1.01\ g/mol$ and O is $16.00\ g/mol$. So, $M_{H_2O}=2\times1.01+16.00 = 18.02\ g/mol$.

Step5: Calculate mass of $H_2O$

The mass of $H_2O$, $m_{H_2O}=n_{H_2O}\times M_{H_2O}=8.1\ mol\times18.02\ g/mol\approx146\ g$.

Answer:

$146\ g$

Question 23

First, balance the equation: $2KClO_3\rightarrow2KCl + 3O_2$.

Explanation:

Step1: Calculate moles of $O_2$

We know that $1\ mol$ of any substance contains $6.022\times10^{23}$ molecules. So, the number of moles of $O_2$, $n_{O_2}=\frac{8.24\times10^{25}}{6.022\times10^{23}}\approx137\ mol$.

Step2: Determine mole - ratio from the balanced equation

The mole - ratio of $KClO_3$ to $O_2$ is $2:3$. So, the number of moles of $KClO_3$, $n_{KClO_3}=\frac{2}{3}n_{O_2}=\frac{2}{3}\times137\ mol\approx91.3\ mol$.

Step3: Calculate molar - mass of $KClO_3$

The molar mass of K is $39.10\ g/mol$, Cl is $35.45\ g/mol$, and O is $16.00\ g/mol$. So, $M_{KClO_3}=39.10+35.45+3\times16.00 = 122.55\ g/mol$.

Step4: Calculate mass of $KClO_3$

The mass of $KClO_3$, $m_{KClO_3}=n_{KClO_3}\times M_{KClO_3}=91.3\ mol\times122.55\ g/mol\approx11200\ g$.

Answer:

$11200\ g$

Question 24

Explanation:

Step1: Calculate molar - mass of $NH_4NO_3$

The molar mass of N is $14.01\ g/mol$, H is $1.01\ g/mol$, and O is $16.00\ g/mol$. So, $M_{NH_4NO_3}=2\times14.01+4\times1.01 + 3\times16.00=80.04\ g/mol$.

Step2: Calculate moles of $NH_4NO_3$

The number of moles of $NH_4NO_3$, $n_{NH_4NO_3}=\frac{m_{NH_4NO_3}}{M_{NH_4NO_3}}=\frac{78.2\ g}{80.04\ g/mol}\approx0.977\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $NH_4NO_3(s)\rightarrow N_2O(g)+2H_2O(l)$, the mole - ratio of $NH_4NO_3$ to $H_2O$ is $1:2$. So, the number of moles of $H_2O$, $n_{H_2O}=2n_{NH_4NO_3}=2\times0.977\ mol = 1.954\ mol$.

Step4: Calculate molar - mass of $H_2O$

The molar mass of H is $1.01\ g/mol$ and O is $16.00\ g/mol$. So, $M_{H_2O}=2\times1.01+16.00 = 18.02\ g/mol$.

Step5: Calculate mass of $H_2O$

The mass of $H_2O$, $m_{H_2O}=n_{H_2O}\times M_{H_2O}=1.954\ mol\times18.02\ g/mol\approx35.2\ g$.

Answer:

$35.2\ g$