Predict the products for nacl+cabr2

Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo

Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess. There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0

3 years ago

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

<u><em>The volume of the gas is 2.80 L.</em></u>

7 0

2 years ago

New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex

marshall27 [118]

Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.$

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

$125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}$

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is: