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3 years ago

How many grams C3H7OH can be made by reacting with 7.3L of CO2 at STP

Chemistry

1 answer:

Komok [63]3 years ago

5 0

Answer:

6.54g of C3H7OH

Explanation:

Step 1:

Determination of the number of mole of CO2 that occupy 7.3L at stp.

This can be obtained as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, Xmol of CO2 will occupy 7.3L at stp i.e

Xmol of CO2 = 7.3/22.4

Xmol of CO2 = 0.326 mole.

Therefore, 0.326 mole of CO2 was used in the reaction.

Step 2:

The balanced equation for the reaction. This is given below:

6CO2 + 8H2O —> 2C3H7OH + 9O2

Step 3:

Determination of the number of mole of C3H7OH produced from the reaction. This is illustrated below:

From the balanced equation above,

6 moles of CO2 reacted to produce 2 moles of C3H7OH.

Therefore, 0.326 mole of CO2 will react to produce = (0.326 x 2)/6 = 0.109 mole of C3H7OH.

Step 4:

Conversion of 0.109 mole of C3H7OH to grams. This is illustrated below:

Number of mole of C3H7OH = 0.109 mole.

Molar mass of C3H7OH = (12x3)+ (7x1) + 16 + 1 = 60g/mol

Mass of C3H7OH =..?

Mass = mole x molar mass

Mass of C3H7OH = 0.109 x 60

Mass of C3H7OH = 6.54g.

Therefore, 6.54g of C3H7OH is produced from the reaction.

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7 0

3 years ago

How do you solve this ??

ICE Princess25 [194]

Answer:

Option D. 400 mmHg

Explanation:

The following data were obtained from the question:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total pressure = 10³ mmHg

Partial pressure of He =.?

Next, we shall determine the total number of mole in the reaction vessel.

This can be obtained as follow:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total mole =?

Total mole = nHe + nNe

Total mole = 0.04 + 0.06

Total mole = 0.1

Next, we shall determine the mole fraction of He.

This can be obtained as follow:

Mole fraction = mole of gas /total mole

Mole of He (nHe) = 0.04 mole

Total mole = 0.1

Mole fraction of He =.?

Mole fraction of He = nHe/total mole

Mole fraction of He = 0.04/0.1

Mole fraction of He = 0.4

Finally, we shall determine the partial pressure of He as follow:

Partial pressure = mole fraction x total pressure

Mole fraction of He = 0.4

Total pressure = 10³ mmHg

Partial pressure of He =.?

Partial pressure of He = 0.4 x 10³

Partial pressure of He = 400 mmHg.

Therefore, the partial pressure of He is 400 mmHg.

8 0

3 years ago

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

(B) is correct.

4 0

3 years ago

Write a net ionic equation that describes a slightly basic solution resulting from dissolving nac2h3o2 in water. Identify all of

Natali [406]

Answer: The net ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + OH^{-}(aq)$

Explanation:

The chemical equation for the given reaction is as follows.

$CH_{3}COONa(aq) + H_{2}O(l) \rightleftarrow CH_{3}COOH(aq) + NaOH(aq)$

We know that a strong acid or base will dissociate completely into a solvent whereas a weak acid or base dissociates partially into the solvent. Hence, the ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + Na^{+}(aq) H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + Na^{+}(aq) + OH^{-}(aq)$ Now, we will cancel the spectator ions from the above equation. Therefore, the net ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + OH^{-}(aq)$

or, $C_{2}H_{3}O^{-}_{2}(aq) + H_{2}O(l) \rightleftharpoons HC_{2}H_{3}O_{2}(aq) + OH^{-}(aq)$

4 0

4 years ago

What is the mass in grams of 0.280 mole sample of sodium hydroxide NaOH

vivado [14]

Answer:

The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.

Explanation:

To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.

So you know:

Na: 23 g/mole
O: 16 g/mole
H: 1 g/mole

So, the molar mass of NaOH is:

NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole

Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?

$mass=\frac{0.28 moles*40 grams}{1 mole}$

mass= 11.2 grams

The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.

4 0

3 years ago