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3 years ago

How much sodium chloride is produced from the synthesis of 100g of sodium and an excess of chlorine gas? 2Na + Cl2= 2NaCl

Chemistry

1 answer:

grigory [225]3 years ago

5 0

Answer:

254.35 g of NaCl.

Explanation:

We'll begin by writing the balanced equation for the reaction between sodium (Na) and chlorine (Cl₂) to produce sodium chloride (NaCl). This is illustrated below:

2Na + Cl₂ —> 2NaCl

Next, we shall determine the mass of Na that reacted and the mass of NaCl produced from the balanced equation. This can be obtained as follow:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 d

Molar mass of NaCl = 23 + 35.5

= 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g

SUMMARY:

From the balanced equation above,

46 g of Na reacted to produce 117 g of NaCl.

Finally, we shall determine the mass of NaCl produced by the reaction of 100 g of Na. This can be obtained as follow:

From the balanced equation above,

46 g of Na reacted to produce 117 g of NaCl.

Therefore, 100 g of Na will react to produce = (100 × 117)/46 = 254.35 g of NaCl.

Thus, 254.35 g of NaCl were obtained from the reaction.

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3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

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