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2 years ago

Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen

Chemistry

1 answer:

Mkey [24]2 years ago

3 0

Answer:

Oxygen is limiting reactant

Explanation:

2 H2 + O2 ======> 2 H2 O

from this equation (and periodic table) you can see that

4 gm of H combine with 32 gm O2

H / O = 4/32 = 1/8

32 /16 = 2/1 shows O is limiter

for 32 gm H you will need 256 gm O and you only have 16 gm

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Please help with this, i don't really get how to do this

LuckyWell [14K]

#4 and #5:
To find pH given concentration of H+ or H30+
pH = - log (H+ or H30+ M)

To find pH given concentration of OH-
Since you already found the pH for this (in #4), you subtract #4's answer from 14.
14 - (pH) = pOH

8 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, $C_6H_4N_2O_4$

3 0

3 years ago

Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y

d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = $10^{-3.5}$ = $3.162 * 10^{-4}$

From the Ice table

$3.162 * 10^{-4}$ = $\frac{x + H^+}{[aspirin]}$ = $\frac{x^{2} }{0.012-x}$

given that the value of Ka is small we will ignore -x

x² = $3.162 * 10^{-4} * 0.012$

x = $1.948 * 10^{-3}$

Therefore

[ H⁺ ] = $1.948 * 10^{-3}$

given that

pH = - Log [ H⁺ ]

= - ( -3 + log 1.948 )

= 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0

2 years ago

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.

Sindrei [870]

Answer:

The percent yield of the reaction is 35 %

Explanation:

In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.

Let's verify the moles that were used in the reaction.

2.05 g . 1mol/ 32 g = 0.0640 mol

In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.

Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).

1atm . 0.550L = n . 0.082 . 295K

(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles

Percent yield of reaction = (Real yield / Theoretical yield) . 100

(0.0225 / 0.0640) . 100 = 35%

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Effectus [21]

Answer:

Mountain ranges are formed by a variety of geological processes, but most of the significant ones on Earth are the result of plate tectonics. Mountain ranges are also found on many planetary mass objects in the Solar System and are likely a feature of most terrestrial planets.

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3 years ago

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