Which represents a balanced nuclear equation?

What mass, in grams, of CO2 and H2O is formed from 2.55 mol of propane?

oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Being the molar mass of each compound:

C₃H₈: 44 g/mole
O₂: 16 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

C₃H₈: 1 mole* 44 g/mole= 44 grams
O₂: 5 moles* 16 g/mole= 80 grams
CO₂: 3 moles* 44 g/mole= 132 grams
H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

$mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}$

mass of CO₂= 336.6 grams

If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

$mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}$

mass of H₂O= 183.6 grams

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

3 0

3 years ago

0.000068 g Express your answer as an integer.

ivann1987 [24]

6.8x10^5? Scientific notation

5 0

3 years ago

Read 2 more answers

WORTH 100 POINTS PLEASE HURRY!!!!!!!!

neonofarm [45]

Answer:

The answer is B

Explanation:

Less air means less air pressure. At a certain point, there is very little air pressure to push against the outward pressure of the helium inside the balloon. It will then expand until the rubber breaks, and the balloon bursts

5 0

3 years ago

Read 2 more answers

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

kaheart [24]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

6 0

2 years ago

If you were to mass the reactants before a reaction and then mass the products after the reaction what would you expect to find

TiliK225 [7]

Answer:

No change...

Explanation:

The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases equals the original mass of the charcoal and the oxygen when it first reacted. So the mass of the product equals the mass of the reactant.

Mark as Brainliest please...

7 0

2 years ago