20) balance the following chemical equation. what number should be on the yellow line? 2 bf3 + 3 li2so3 -->…

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\rightarrow1B_2(SO_3)_3+6LiF).

Answer:

1

Explanation 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\rightarrow2AlCl_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 for sub - question 21:

43.6 g