Question

1. Calculate the number of AlCl3 that are produced from 0.25 mol of Al(OH)3.
Al(OH)3 + 3HCl --> 3H2O+AlCl3
molar mass of AlCl3 = 133.3g
a. 10.0g
b. 33.3g
c. 8.25g
d. 20.0g

2. Given the balanced equation 4NH3 + 5O2 --> 4NO + 6H2O
How many grams of NO are produced when 12 moles of O2 are consumed?
molar mass NO= 30.0 g
a. 288g
b. 384g
c. 450g
d. 30.0g

Answer 1

Answer:

$1.(b)33.3g \\ 2.(a)288g$

Explanation:

1.The correct answer is (b) 33.3g.

To calculate the number of AlCl3 produced, you can use the stoichiometry of the reaction:

Al(OH)3 + 3HCl → 3H2O + AlCl3

From the balanced equation, you can see that 1 mole of Al(OH)3 produces 1 mole of AlCl3. Since you are given 0.25 mol of Al(OH)3, you can calculate the number of moles of AlCl3 produced:

0.25 mol Al(OH)3 × 1 mol AlCl3/1 mol Al(OH)3 = 0.25 mol AlCl3

Finally, you can convert the number of moles of AlCl3 to grams by multiplying by the molar mass of AlCl3 (133.3 g/mol):

0.25 mol AlCl3 × 133.3 g/mol = 33.3 g AlCl3

Therefore, the correct answer is (b) 33.3g.

2. The correct answer is (a) 288g

First, use the balanced equation to determine the number of moles of NO produced:

4NH3 + 5O2 → 4NO + 6H2O

From the equation, 5 moles of O2 produce 4 moles of NO. Since 12 moles of O2 are consumed, you can calculate the number of moles of NO produced:

12 mol O2 × 4 mol NO/5 mol O2 = 9.6 mol NO

Next, multiply the number of moles of NO by the molar mass of NO (30.0 g/mol) to get the mass of NO produced:

9.6 mol NO × 30.0 g/mol = 384 g NO

Therefore, the correct answer is (a)288g.

Answer 2

Answer:

1) b. 33.3 g

2) a. 288g

Explanation:

Part 1:

To solve this problem, we need to first determine the stoichiometry of the reaction, and then use the given information to find the number of moles of $\sf AlCl_3$ produced.

The balanced chemical equation for the reaction is:

$\textsf{Al(OH)}_3 + 3\textsf{HCl} \longrightarrow 3\textsf{H}_2\text{O} + \textsf{AlCl}_3$

From the balanced equation, we can see that 1 mole of $\sf Al(OH)_3$ produces 1 mole of $\sf AlCl_3$.

Given that 0.25 mol of $\sf Al(OH)_3$ is reacted, it will produce 0.25 mol of $\sf AlCl_3$.

Now, to find the mass of $\sf AlCl_3$ produced, we'll use the molar mass of $\sf AlCl_3$, which is 133.3 g/mol:

$\textsf{Mass of AlCl}_3 = \text{Number of moles} \times \text{Molar mass}$

$\textsf{Mass of AlCl}_3 = 0.25 \, \text{mol} \times 133.3 \, \text{g/mol}$

$\textsf{Mass of AlCl}_3 = 33.325 \, \text{g}$

Therefore, the correct answer is option b. 33.3g.

$\hrulefill$

Part 2:

To determine the amount of NO produced when 12 moles of Oxygen are consumed, we can use the stoichiometry of the balanced chemical equation. The given balanced equation is:

$4 \textsf{NH}_3 + 5 \textsf{O}_2 \longrightarrow 4 \textsf{NO} + 6 \textsf{H}_2\textsf{O}$

The coefficients in the balanced equation tell us the mole ratio between reactants and products. In this case, the coefficient of $\sf O_2$ is 5, which means 5 moles of $\sf O_2$ react to produce 4 moles of NO.

Given that 12 moles of $\sf O_2$ are consumed, we can calculate the moles of NO produced:

$\textsf{Moles of NO} = \dfrac{4}{5} \times 12\\\\ = 9.6 \text{ moles}$

Now, we can convert moles of NO to grams using the molar mass of NO:

$\textsf{Mass of NO} = \textsf{Moles of NO} \times \textsf{Molar Mass of NO}$

$\textsf{Mass of NO} = 9.6 \, \textsf{moles} \times 30.0 \, \textsf{g/mol}$

$\textsf{Mass of NO} = 288.0 \, \textsf{g}$

Therefore, the correct answer is option a. 288g.