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Dmitrij [34]
2 years ago
9

The chemical equation below shows the burning of magnesium (Mg) with oxygen (O2) to form magnesium oxide (MgO). 2Mg + O2 Right a

rrow. 2MgO The molar mass of O2 is 32.0 g/mol. What mass, in grams, of O2 is required to react completely with 4.00 mol of Mg? 2.00 64.0 128 256
Chemistry
1 answer:
Nadusha1986 [10]2 years ago
4 0

Answer:

64.0

Explanation:

2Mg+O2 ---> 2MgO

use dimentional analysis to find the amount of moles of O2 needed first

4.00molMg x 1.00mol O2/ 2.00 mol Mg=. 2.00 mol O2

using the coefficients you can see the mole ratio for O2:Mg the mole ratio is 1:2 which is why there is 1 mole on the top for 2 moles on the bottom. The Mg would cancel and multiply 4 by 1 then divide by 2, or multipy 4 by 1/2

Now that you have the moles of O2 you use the molar mass to find the grams in 2 moles of O2

2.00 mol O2 x 32.0g/1.00 mol = 64.0 g

multiply 2 by 32

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7 0
2 years ago
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
2 years ago
A solution was prepared by mixing 50.0 g
frez [133]

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

4 0
2 years ago
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