Ask question

Ask question

All categories

2 years ago

7

When 189.6 g of ethylene (C2H4) burns in oxygen to give carbon dioxide and water, how many grams of CO2 are formed? C2H4(g) + O2

(g) → CO2(g) + H2O(g) (unbalanced)

1 answer:

ser-zykov [4K]2 years ago

6 0

Answer:

596 g of CO₂ is the mass formed

Explanation:

Combustion reaction:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

We determine moles of ethylene that has reacted:

189.6 g . 1mol / 28g = 6.77 moles

We assume the oxygen is in excess so the limiting reagent will be the ethylene.

1 mol of ethylene produce 2 moles of CO₂ then,

6.77 moles will produce the double of CO₂, 13.5 moles.

We convert the moles to mass: 13.5 mol . 44 g /1mol = 596 g

You might be interested in

Valence bond theory predicts that bromine will use _____ hybrid orbitals in brf5.

koban [17]

Answer:

the correct answer is (sp3d2) (d)

Explanation:

5 0

2 years ago

If a typical antacid tablet contains 2.0 g of sodium hydrogen carbonate, how many moles of carbon dioxide should one tablet yiel

ivolga24 [154]

The equation of the chemical reaction is NaHCO3 + H+ --> H2O + CO2 + Na
To determine the total number of moles of carbon dioxide, the given mass of sodium hydrogen carbonate is divided with its own molar mass. Then it is multiplied with the ratio between NaHCO3 and carbon dioxide. The total number of moles of CO2 one tablet should yield is 0.024 mole.

6 0

2 years ago

Read 2 more answers

The pressure and temperature inside a bike tire is 10 atm and 10k respectively. What will the pressure become in the tire when t

igomit [66]

Answer:

$P_2=20atm$

Explanation:

Hello,

In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:

$\frac{P_1}{T_1}= \frac{P_2}{T_2}$

Thus, we solve for the final pressure P2 to obtain it as shown below:

$P_2=\frac{P_1T_2}{T_1}=\frac{10atm*20K}{10K} \\\\P_2=20atm$

Hence, we notice that the temperature doubles as well as the pressure.

Best regards.

3 0

2 years ago

If the pH of a phosphoric acid (H3PO4) solution is adjusted to 6.50, what is the most abundant species and which is the second m

maria [59]

Explanation:

For the given values of $K_{a}$ we will have the values of $pK$ as follows.

As, $pK_{a} = -log K_{a}$

Therefore,

$pK_{a1}$ = 2.15, $pK_{a2}$ = 7.20

$pK_{a3}$ = 12.38

Now, at pH 6.50

$H_{3}PO_{4} \rightarrow H_{2}PO^{-}_{4} + H^{+}$ ; $K_{a1}$

At pH = 2.15; $H_{3}PO_{4} = H_{2}PO^{-}_{4}$

$H_{2}PO^{-}_{4} \rightarrow HPO^{2-}_{4} + H^{+}$ ; $K_{a2}$

At pH 7.20; $H_{2}PO^{-}_{4} = HPO^{2-}_{4}$

$HPO^{2-}_{4} \rightarrow PO^{3-}_{4} + H^{+}$ ; $K_{a3}$

Hence, we can conclude that most abundant species is $H_{2}PO^{-}_{4}$ and the second most abundant species is $HPO^{2-}_{4}$ .

3 0

2 years ago

A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in

Fittoniya [83]

The molarity (M) of the acid in the lake water is $0.00001M$ .

<u>Explanation:</u>

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

$M_1V_1=M_2V_2$

Here;

$M_1,M_2$ are molarity of acid in the lake water and base solution respectively. $V_1,V_2$ are volume of sample in the lake water and base solution respectively.

Given values are as follows:

$M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml$

Putting these values in above equation :

$M_1V_1=M_2V_2$

$M_1(1000)=(0.100)(0.100)$

$M_1=\frac{(0.100)(0.100)}{1000}$

$M_1=0.00001M$

Therefore, the molarity (M) of the acid in the lake water is $0.00001M$ .

5 0

1 year ago