All categories

2 years ago

What is the first thing you must do to solve a stoichiometry problem

Chemistry

1 answer:

belka [17]2 years ago

3 0

Answer:

the first step in any stoichiometric problem is to always ensure that the chemical reaction you are dealing with is balanced,

Explanation:

You might be interested in

The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

<u>Answer:</u> The moles of $CO_2$ added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago

A food substance kept at 0Â°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6

ZanzabumX [31]

Answer:

1. 67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

$In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

Given that:

time $t_1$ = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time $t_2$ = 10.6 hours

Temperature $T_1$ = 0° C = (0+273 )K = 273 K

Temperature $T_2$ = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since $(\frac{k_2}{k_1}=\frac{t_2}{t_1})$

Then we can rewrite the above expression as:

$In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

$In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})$

$2.934 = \frac{E_a}{8.314}(\frac{30}{82719})$

$2.934 = \frac{30E_a}{687725.766}$

$30E_a = 2.934 *687725.766$

$E_a = \frac{2.934 *687725.766}{30}$

$E_a =67255.58 \ J/mol$

$E_a =67.2 \ kJ/mol$

7 0

3 years ago

What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso

Paraphin [41]

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

6 0

3 years ago

is trying to plate gold onto his silver ring. He constructs an electrolytic cell using his ring as one of the electrodes. He run

notka56 [123]

Answer:

0.013 mole.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 94.6 mins

Current (I) = 224.8 mA.

Mole of electrons (e) =?

Next, we shall convert 94.6 mins to seconds. This is illustrated below:

1 min = 60 s

Therefore,

94.6 mins = 94.6 min × 60 s / 1 min

94.6 mins = 5676 s

Thus, 94.6 mins is equivalent to 5676 s.

Next, we shall convert 224.8 mA to ampere (A). This can be obtained as follow:

1 mA = 1×10¯³ A

Therefore,

224.8 mA = 224.8 mA × 1×10¯³ A / 1 mA

224.8 mA = 0.2248 A

Thus, 224.8 mA is equivalent to 0.2248 A.

Next, we shall determine the quantity of electricity flowing in circuit. This can be obtained as follow:

Time (t) = 5676 s

Current (I) = 0.2248 A

Quantity of electricity (Q) =?

Q = it

Q = 0.2248 × 5676

Q = 1275.96 C

Finally, we shall determine the mole of electron by converting the quantity of electricity (i.e 1275.96 C) to number of electrons. This can be obtained as follow:

96500 C = 1 e

Therefore,

1275.96 C = 1275.96 C × 1 e / 96500 C

96500 C = 0.013 e

Thus, the mole of electron trasfered in the process is 0.013 mole.

3 0

3 years ago

Which of the following statements is true?

Valentin [98]

d is the correct answer

4 0

4 years ago

Read 2 more answers