What is the amount of aluminum chloride produced from 40 moles of aluminum and excess chlorine?

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+

Vera_Pavlovna [14]

Answer:

5 moles of NO₂ will remain after the reaction is complete

Explanation:

We state the reaction:

3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)

3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:

If 3 moles of nitric oxide need 1 mol of water to react

Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O

We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:

1 mol of water needs 3 moles of oxide to react

Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide

We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains

5 0

3 years ago

What are differences and similarities between pure substances and mixtures

Snezhnost [94]

Answer:

HELLO THERE!

I HOPE MY ANSWER WILL HELP YOU :)

Explanation:

There is this picture that helped me. I hope it helps you too.

5 0

3 years ago

A 951g sample of carbon steel with a specific heat capacity of .49j is heated to 673k and quenched in oil. The steel cools to 52

nikdorinn [45]

Answer: all I know it’s not -31.5 for ppl taking the k12 test

Explanation: I took the test

5 0

3 years ago

Read 2 more answers

Is a oreo a pure substance or a mixture

adoni [48]

Belive it or not but an oreo is a mixture. LOL

7 0

3 years ago

Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →

Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = $\frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }$

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0

3 years ago