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GenaCL600 [577]

2 years ago

11

When calcium is allowed to react with nitrogen, calcium nitride is formed. If 24.0 g of calcium and 12.0 g of nitrogen are avail

able for this reaction, the limiting reagent will be?

1 answer:

Thepotemich [5.8K]2 years ago

8 0

What are the mol ratios? if you’re just comparing with the mass (which i doubt so since most limiting questions require u to compare the mols)

but if u are, the limiting would be nitrogen since it’s lesser than calcium mass

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Balance the equation for ethane C3H6 burning in oxygen to form carbon dioxide and steam. _____C3H6 + ____ O2 ---> CO2 + ____H

aleksandrvk [35]

Answer:

2C3H6 + 9 O2 ---> 6 CO2 + 6 H2O

Explanation:

8 0

2 years ago

The bonds in the compound MgSO4 can be described as

Butoxors [25]

C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond

7 0

3 years ago

(58 g)/ (4L) reduce units to one

katen-ka-za [31]

Answer is 14.5 g L⁻¹.

<em>Explanation;</em>

Here, the question says reduce the units as one.

The presented units are g/L. To reduce the units to one, what we can do is take L to the upper side.

This can be done according to the rules of indices;

1 / aˣ = a⁻ˣ

Like that, we can write 1 / L as L⁻¹.

Hence, the reduced unit is g L⁻¹.

But remember to keep a space between when writing two different units.

Actually, this is an unit for density.

3 0

3 years ago

How does the sun heat<br> the side of the Earth<br> facing away from it?

Evgen [1.6K]

Answer:

The sun radiates energy in all directions.

Explanation:

Most of it dissipates into space, but the tiny fraction of the sun's energy that reaches Earth is enough to heat the planet and drive the global weather system by warming the atmosphere and oceans

3 0

3 years ago