Question

1. balance the following chemical equation. what number should be on the yellow line? 4nh₃ + 5o₂ -> 4no + yellow lineh₂o
2. balance the following chemical equation. what number should be on the yellow line? 2bf₃ + 3li₂so₃ -> yellow lineb₂(so₃)₃ + 6lif
3. if 8.82 grams of aluminum reacts, how many grams of alcl₃ will be produced? 2al + 6hcl -> 2alcl₃ + 3h₂

chemical reactions & evidence of chemical reaction

Explanation:

Step1: Balance hydrogen atoms in the first equation

In the equation $4NH_3 + 5O_2
ightarrow4NO+\underline{?}H_2O$, on the left - hand side, the number of hydrogen atoms is $4\times3 = 12$. On the right - hand side, in $H_2O$, if we let the coefficient be $x$, then the number of hydrogen atoms is $2x$. So, $2x = 12$, and $x = 6$.

Step2: Balance boron atoms in the second equation

In the equation $2BF_3+3Li_2SO_3
ightarrow\underline{?}B_2(SO_3)_3 + 6LiF$, on the left - hand side, the number of boron atoms is 2. On the right - hand side, in $B_2(SO_3)_3$, if we let the coefficient be $y$, then the number of boron atoms is $2y$. So, $2y=2$, and $y = 1$.

Step3: Calculate the mass of $AlCl_3$ produced in the third equation

1. First, find the molar mass of $Al$ and $AlCl_3$.

• The molar mass of $Al$ ($M_{Al}$) is approximately $26.98\ g/mol$, and the molar mass of $AlCl_3$ ($M_{AlCl_3}$) is $26.98+3\times35.45=26.98 + 106.35=133.33\ g/mol$.
• The number of moles of $Al$ ($n_{Al}$) in $8.82\ g$ of $Al$ is $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{8.82\ g}{26.98\ g/mol}\approx0.327\ mol$.
• From the balanced equation $2Al + 6HCl

ightarrow2AlCl_3+3H_2$, the mole ratio of $Al$ to $AlCl_3$ is $1:1$. So, the number of moles of $AlCl_3$ produced ($n_{AlCl_3}$) is equal to the number of moles of $Al$ reacted, $n_{AlCl_3}=0.327\ mol$.

• The mass of $AlCl_3$ produced ($m_{AlCl_3}$) is $m_{AlCl_3}=n_{AlCl_3}\times M_{AlCl_3}=0.327\ mol\times133.33\ g/mol\approx43.6\ g$.

Answer:

1. 6
2. 1
3. 43.6 g