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3 years ago

IF 14.07*10^26 molecules of magnesium chloride was produced in the following reaction, how many grams of magnesium reacted?

Chemistry

1 answer:

Wittaler [7]3 years ago

6 0

Answer:

56160grams

Explanation:

First, we need to convert the number of molecules of magnesium chloride (MgCl2) into moles by dividing by Avagadro's number (6.02 × 10^23 molecules)

n = nA ÷ 6.02 × 10^23

n = 14.07 × 10^26 ÷ 6.02 × 10^23

n = 14.07/6.02 × 10^(26-23)

n = 2.34 × 10^3 moles of MgCl2

The balanced reaction given in the question is as follows:

Mg + 2HCl → MgCl2 + H2

If 1 mole of Mg produced 1 mole of MgCl2

Then, 2.34 × 10^3 moles of Mg will also produce 2.34 × 10^3 moles of MgCl2.

Using mole = mass ÷ molar mass (MM)

Molar mass of Mg = 24g/mol

mass = mole × MM

mass = 2.34 × 10^3 × 24

mass = 56.16 × 10^3

mass = 56160grams.

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Pavlova-9 [17]

Reaction is NaHCO3 + HCL = H2CO3 + NaCl
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the 8 moles of HCl are in excess

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3 years ago

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Some commercially available algaecides for swimming pools claim to contain 7% copper. Could the method used in this experiment t

AleksandrR [38]

Answer:

Explanation:

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4 0

4 years ago

How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has

Papessa [141]

The balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas= $\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.

number of moles of sodium azide = $16 moles$

Mass of sodium azide in g = $number of moles\times molar mass of sodium azide$ .

= $16 moles\times 65.00 g/mol$

= $1040 g$

Thus, mass of sodium azide which is required to produce $19.0 ft^{3}$ of nitrogen gas = $1040 g$

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3 years ago

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Evgen [1.6K]

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2 years ago