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3 years ago

How do you find the limiting reactant if 2.50 mol copper and 5.50 mol silver nitrate react by single displacement?

Chemistry

1 answer:

Crank3 years ago

6 0

The limiting reagent is Copper (Cu).

Explanation:

To identify the limiting reactant., first balance the reaction.

The balanced equation is:

Cu + 2AgNO3 -----> Cu(NO3)2 + 2Ag

To find the limiting reactant, take the amount of initial substance and find the number of moles of one of the products. The reactant that gives a product with the least number of moles, is the limiting reactant

$5.50 mol Cu \times (\frac{1.00 mol Cu}{2.00 mol AgNO3} )$ = 2.75 mol Cu.

Since there are only 2.50 mol Cu, copper is the limiting reactant, because, with that quantity of copper, only 2.50 mol AgNO3 will be reacted.

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Given 12 g of aluminum, how many moles do you have?

MaRussiya [10]

Answer:

$0.168631 \ moles \ Al$

Explanation:

Atomic mass of $Al$ is given as $26.982u$ , where $u$ is the unified atomic mass unit.

$u$ is defined as $\frac{1}{12}$ th the mass of a neutral carbon-12 atom. This is equal to the mass of $one \ nucleon$ . Therefore:-

$1u=1g/mol\\$

Mole constant= $6.022\times10^{23}$ atoms of any element.

For $Al$ of 12g:-

$12\times \frac{1 \ mole \ Al}{26.982g}\\=0.168631 \ moles$ $=12\times\frac{1 \ mole \ Al}{26.982g}\\=0.168631 \ moles \ Al$

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3 years ago

A ______________ is a prediction of an outcome and the basis for experimentation.

a_sh-v [17]

A hypothesis is a prediction of an outcome and the basis for experimentation.

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4 years ago

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3. Explain why environmental science is an interdisciplinary science.

weqwewe [10]

Answer:

Environmental science is also referred to as an interdisciplinary field because it incorporates information and ideas from multiple disciplines. Within the natural sciences, such fields as biology, chemistry, and geology are included in environmental science.

Explanation:

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4 years ago

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The molar mass of a certain gas is 49 g. What is the density of the gas in g/L at STP?

snow_tiger [21]

Answer:

$\boxed{\text{2.2 g/L}}$

Explanation:

We can use the Ideal Gas Law to calculate the density of the gas.

pV = nRT

n = m/M Substitute for n

pV = (m/M)RT Multiply both sides by M

pVM = mRT Divide both sides by V

pM = (m/V) RT

ρ = m/V Substitute for m/V

pM = ρRT Divide each side by RT

$\rho = \frac{pM }{RT}$

Data:

p = 1.00 bar

M = 49 g/mol

R = 0.083 14 bar·L·K⁻¹mol⁻¹

T = 0 °C = 273.15 K

Calculation:

ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L

The density of the gas is $\boxed{\text{2.2 g/L}}$ .

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4 years ago

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

stiv31 [10]

Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's y = C

y = 3x

Still not converted sum of reaction

for A: 100 - 2x

for B: 50 - x

Shift of x over time

$\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}$

Integration of x as regards t

$\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\ \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\$

D is the constant of integration

initial conditions: t = 0, x = 0

$\frac{-1}{[100-2x]} = -kt + D \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\$

hence we get:

$\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}$

after t = 7 minutes , $C = 10 \ g = 3x$

$3x = 10\\\\x = \frac{10}{3}$

Insert the above value x into $\frac{1}{(100-2x)}$ equation $= kt + \frac{1}{100}$ to get k.

$\to \frac{1}{(100-2\times \frac{10}{3})} = k \times (7) + \frac{1}{100} \\\\ \to \frac{1}{(100- 2 \times 3.33)} = \frac{700k + 1}{100} \\\\ \to \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\$

$\to 100 = 93.34(700k + 1) \\\\ \to 100 = 65,338k + 700 \\\\ \to 65,338k = -600 \\\\ \to k = \frac{-600}{ 65,338} \\\\ \to k= - 0.0091$

therefore plugging in the equation the above value of k

$\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} = \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}$