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Vlad [161]
4 years ago
8

Hydrogen reacts with sodium to produce solid sodium hydride. A reaction mixture contains 6.75 g Na and 3.03 g Hydrogen. Which re

actant is limiting?
Chemistry
1 answer:
Colt1911 [192]4 years ago
6 0

Answer:

Na

Explanation:

To find the limiting reactant we must first write the balanced chemical reaction. It must be correctly balanced so that we can find the proper mole ratios.

2 Na + H₂ → 2 NaH

After this we will convert our measurements to moles. For mass we do this by dividing by the molar mass.

6.75g ÷ 22.99 = 0.294mol Na

3.03g ÷ 2.02 = 1.50mol H₂

Now that we have the moles of each of the reactants, we can multiply them by their mole ratio with a reactant.

0.294mol Na × \frac{2 Na}{2 NaH} = 0.294mol NaH

1.50mol H₂ × 2/1 = 3.00mol NaH

Na is our limiting reagent because it makes the smaller amount of moles.

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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.
leonid [27]
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.

31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag         6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻   →   ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻  = 107.9 g Ag
       ∅        | 63.5 g Cu | 1 mol Cu | 1 mol Ag              63.5


There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
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3 years ago
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Why does pepsinogen become activated at ph 2?
Katena32 [7]
////// it is pepsinogen //////
5 0
3 years ago
HELP ASAP MY LESSON IS BEING TIMED!!!
77julia77 [94]

Answer:

1. Newton's third law of motion

2. Unbalanced

3. Action

7 0
3 years ago
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A 5-kg quantity of radioactive isotope decays to 2 kg after 10 years. Find the decay constant of the isotope.k = _____
Dvinal [7]

Answer:

k = -0.09165 years^(-1)

Explanation:

The exponential decay model of a radioactive isotope is generally given as;

A(t) = A_o(e^(kt))

Where;

A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years

We are given;

A_o = 5 kg

A(10) = 2kg

t = 10 years

Thus;

A(10) = 2 = 5(e^(10k))

Thus;

2 = 5(e^(10k))

2/5 = (e^(10k))

0.4 = (e^(10k))

In 0.4 = 10k

-0.9164 = 10k

k = -0.9164/10

k = -0.09165 years^(-1)

7 0
3 years ago
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