All categories

3 years ago

How many moles of hydrogen, H2, are needed to react with 4.0 moles of nitrogen, N2

1 answer:

3 0

6 moles because you start with a balanced equation and with the balanced equation it means that regardless of how many miles of nitrogen gas you have, the reaction will always consume twice as many miles if hydrogen gas.

You might be interested in

How many protons does the largest transition metal have in its nucleus?

CaHeK987 [17]

The largest transition metal is copernicium with 112 protons

4 0

3 years ago

Read 2 more answers

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci

Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

8 0

3 years ago

What is the atomic number of the yet-undiscovered element directly below francium ( Fr) in the periodic table?

kakasveta [241]

The atomic number of Francium is 87.

4 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

Determine the LIMITING reactant in the following balanced equation:

padilas [110]

Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂ → 2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

KBr : KCl

2 : 2

0.03 : 0.03

KBr : Br₂

2 : 1

0.03 : 1/2×0.03= 0.015

Cl₂ : KCl

1 : 2

0.09 : 2/1×0.09 = 0.18

Cl₂ : Br₂

1 : 1

0.09 : 0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.

5 0

3 years ago