Oxygen gas and water are produced by the decomposition of hydrogen peroxide. If 10.0 mol of H2O2 decomposes, what volume of oxygen will be produced? Assume the density of oxygen is 1.429g/L. 2H2O2= 2H2O+O2

First, we need to determine the moles of O2 produced from the decomposition of hydrogen peroxide. From the balanced chemical equation, we can see that for every 2 moles of H2O2, 1 mole of O2 is produced.

2H2O2 -> 2H2O + O2

Therefore, if 10.0 moles of H2O2 decompose, we can use the mole ratio to determine the moles of O2 produced:

10.0 moles H2O2 × (1 mole O2 / 2 moles H2O2) = 5.0 moles O2

Next, we will find the mass of oxygen produced using the molar mass of oxygen, O2 (molar mass of O = 16.00 g/mol; molar mass of O2 = 2 × 16.00 g/mol = 32.00 g/mol):

5.0 moles O2 × 32.00 g/mol = 160.0 g O2

Now, we can use the density of oxygen to find the volume of oxygen produced:

Density = Mass / Volume

1.429 g/L = 160.0 g / Volume

Volume = 160.0 g / 1.429 g/L = 111.9 L

Therefore, if 10.0 mol of H2O2 decomposes, it will produce approximately 111.9 L of oxygen gas.

To find the volume of oxygen produced, you need to follow these steps:

1. Convert the number of moles of hydrogen peroxide (H2O2) into moles of oxygen (O2) using the balanced equation. According to the equation, for every 2 moles of H2O2, 1 mole of O2 is produced. So, if you have 10.0 moles of H2O2, you will produce 10.0/2 = 5.0 moles of O2.

2. Convert the moles of O2 to grams using the molar mass of oxygen, which is 32.00 g/mol. Multiply the number of moles of O2 by the molar mass: 5.0 moles O2 x 32.00 g/mol = 160.0 grams O2.

3. Convert the mass of O2 to volume using its density. Density is defined as mass divided by volume. Rearranging the equation, we get volume = mass/density. Substitute the mass of O2 (160.0 grams) and the density of O2 (1.429 g/L) into the equation:

volume = 160.0 g / 1.429 g/L = 111.8 L

Therefore, when 10.0 mol of H2O2 decomposes, it produces 111.8 liters of oxygen gas.

To determine the volume of oxygen gas produced, we first need to find the moles of oxygen produced.

From the equation 2H2O2 = 2H2O + O2, we can see that the molar ratio of H2O2 to O2 is 2:1.

So for every 2 moles of H2O2 that decompose, 1 mole of O2 is produced.

Given that 10.0 moles of H2O2 decompose, we can calculate the moles of O2 produced:

Moles of O2 = (10.0 moles H2O2) / (2 moles H2O2/1 mole O2)
= 5.0 moles O2

Now, to find the volume of oxygen gas, we can use the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature (assume constant at room temperature: 298 K).

Rearranging the equation, we have:

V = (nRT) / P

Given:
n = 5.0 moles
R = 0.0821 L·atm/(mol·K)
P = density of oxygen gas = 1.429 g/L

First, convert the density of oxygen gas from g/L to mol/L:

1.429 g/L x (1 mol O2 / 32g O2) = 0.0447 mol/L

Now, substitute the values into the equation:

V = (5.0 moles) x (0.0821 L·atm/(mol·K)) x (298 K) / (0.0447 mol/L)
V = 88.0 L

Therefore, the volume of oxygen gas produced from the decomposition of 10.0 mol of H2O2 is 88.0 liters.