Question

1. balance the following chemical equation. what number should be on the yellow line? 2 bf3 + 3 li2so3 --> b2(so3)3 + 6 lif
2. if 8.82 grams of aluminum reacts, how many grams of alcl3 will be produced? 2 al + 6 hcl --> 2 alcl3 + 3 h2

Explanation:

Step1: Analyze boron atoms

On the left - hand side, in \(2BF_3\), the number of boron (B) atoms is 2. On the right - hand side, in \(B_2(SO_3)_3\), the number of boron atoms is 2 when the coefficient of \(B_2(SO_3)_3\) is 1.

Step2: Analyze sulfur and oxygen atoms

In \(3Li_2SO_3\), there are 3 sulfur (S) and 9 oxygen (O) atoms. In \(B_2(SO_3)_3\), when the coefficient is 1, there are 3 sulfur and 9 oxygen atoms.

Step3: Analyze lithium and fluorine atoms

In \(3Li_2SO_3\), there are 6 lithium (Li) atoms, and in \(6LiF\), there are 6 lithium atoms. In \(2BF_3\), there are 6 fluorine (F) atoms, and in \(6LiF\), there are 6 fluorine atoms. So the balanced equation is \(2BF_3 + 3Li_2SO_3
ightarrow1B_2(SO_3)_3+6LiF\).

for sub - question 21:

Step1: Calculate the molar mass of Al

The molar mass of Al (\(M_{Al}\)) is approximately \(26.98\ g/mol\).

Step2: Calculate the number of moles of Al

The number of moles of Al (\(n_{Al}\)) is calculated using the formula \(n=\frac{m}{M}\), where \(m = 8.82\ g\) and \(M = 26.98\ g/mol\). So \(n_{Al}=\frac{8.82\ g}{26.98\ g/mol}\approx0.327\ mol\).

Step3: Determine the mole ratio

From the balanced chemical 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, \(n_{AlCl_3}= 0.327\ mol\).

Step4: Calculate the molar mass of \(AlCl_3\)

The molar mass of \(AlCl_3\) (\(M_{AlCl_3}\)) is \(M_{Al}+3\times M_{Cl}\), where \(M_{Al}=26.98\ g/mol\) and \(M_{Cl} = 35.45\ g/mol\). So \(M_{AlCl_3}=26.98\ g/mol+3\times35.45\ g/mol=26.98\ g/mol + 106.35\ g/mol=133.33\ g/mol\).

Step5: Calculate the mass of \(AlCl_3\)

The mass of \(AlCl_3\) (\(m_{AlCl_3}\)) is calculated using the formula \(m = n\times M\), so \(m_{AlCl_3}=0.327\ mol\times133.33\ g/mol\approx43.6\ g\).

Answer:

1