Ask question

Ask question

All categories

3 years ago

10

Two moles of magnesium and five moles of oxygen are placed in a vessel. When magnesium is ignited ca two moles of magnesium and

five moles of oxygen are placed in a vessel. When magnesium is ignited, if reacts with oxygen. What is the limiting reactant in this experiment

1 answer:

Greeley [361]3 years ago

4 0

Limiting reactant in this experiment would be Magnesium since it will run out first

You might be interested in

How many grams of NaOH are produced from 20.0 grams of Na2CO3?

natita [175]

Answer:

Hope this helps!

Explanation:

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

4 0

3 years ago

Which of the following solutes has the greatest effect on the colligative properties for a given mass of pure water? Explain.

Strike441 [17]

Answer:

Option a.

0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i

Explanation:

To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)

These are the colligative properties:

ΔP = P° . Xm . i → Lowering vapor pressure

ΔT = Kb . m . i → Boiling point elevation

ΔT = Kf . m . i → Freezing point depression

π = M . R . T → Osmotic pressure

Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.

CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:

CaCl₂ → Ca²⁺ + 2Cl⁻

We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3

KNO₃ → K⁺ + NO₃⁻

We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2

Option a, is the best.

8 0

3 years ago

A flask contains 0.83mol of neon gas at a temperature of 35°C. The pressure gauge indicates 0.37atm inside the flask. What is th

Angelina_Jolie [31]

Answer: 5.66 dm3

Explanation:

Given that:

Volume of neon gas = ?

Temperature T = 35°C

Convert Celsius to Kelvin

(35°C + 273 = 308K)

Pressure P = 0.37 atm

Number of moles N = 0.83 moles

Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.37atm x V = 0.83 moles x 0.0082 atm dm3 K-1 mol-1 x 308K

0.37 atm x V = 2.096 atm dm3

V = (2.096 atm dm3 / 0.37atm)

V = 5.66 dm3

Thus, the volume of the neon gas is 5.66 dm3

5 0

3 years ago

A scientist is trying to determine whether a certain substance is radioactive. Which question would be most helpful to ask?

oksian1 [2.3K]

Well a question to ask would be if the mass of the material has changed significantly as that would determine that the substance is radioactive or if there have been any high readings found by a Geiger meter in certain period of time

hope that helps

3 0

4 years ago

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.

padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

#SPJ4

5 0

1 year ago