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3 years ago

How does the law of conservation of mass apply to this reaction:Mg + 2HCl → H2 + MgCl2?

1 answer:

5 0

The law of conservation of mass applies to every reaction. In this case, you start with 1 Mg, 2 H, and 2CL and end up with the same five only their bonds have been rearranged, or in other words, they are joined up differently.

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Calculate the mass of hydrogen in 500 g of Al(C2H3O2)3

Zigmanuir [339]

Answer is: the mass of hydrogen is 22,05 grams.
m(Al(C₂H₃O₂)₃) = 500 g.
M(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.
n(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.

3 0

3 years ago

Read 2 more answers

In what type of reaction will an acid and a base react with each other

Arada [10]

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acidcombines with the OH(-) anion of the base to form water.

6 0

3 years ago

Read 2 more answers

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

3 years ago

Why does a microscope stage have a small hole in it?

jarptica [38.1K]

Answer: B. Allow light to pass through. :)

4 0

3 years ago

Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de car

jarptica [38.1K]

Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.

Usando la ley general de los gases, las moles de dioxido de carbono son:

PV = nRT.

Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = 84.7g de CaCO₃

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100 =

7 0

3 years ago