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2 years ago

Given the equation: 2H2 + O2 → 2H2O, how many moles of oxygen are needed to react with 13 moles of hydrogen

Chemistry

1 answer:

bulgar [2K]2 years ago

6 0

Explanation:

2H2 + O2 = 2H2O

2mol. 1mol. 2mol

2mol reacts with 1mol

13mol reacts with x

x=13mol × 1mol

2mol

x= 13mol

2mol

x= 6.5mol of oxygen

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

6 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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