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2 years ago

2Na + 2H2O → 2NaOH + H2

Chemistry

2 answers:

Zarrin [17]2 years ago

5 0

The initial quantity of sodium metal is 17.25

Alinara [238K]2 years ago

5 0

Answer: 13 grams

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to stoichiometry,

1 mole of $H_2$ is produced from 2 moles of sodium

or 22.4 L of $H_2$ at STP is produced from =2 moles of sodium

thus 6.30 L of $H_2$ at STP is produced from = $\frac{2}{22.4}\times 6.30=0.56moles$ of sodium

To calculate the mass of sodium metal used

$\text{mass of sodium used}=moles\times {\text {Molar mass}}=0.56\times 23=13grams$

Thus the initial quantity of sodium metal used is 13 grams.

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Answer: The empirical formula of the compound is $Fe_1S_1O_4$

Explanation:

Empirical formula is defined formula which is simplest integer ratio of number of atoms of different elements present in the compound.

Percentage of iron in a compound = 36.76 %

Percentage of sulfur in a compound = 21.11 %

Percentage of oxygen in a compound = 42.13 %

Consider in 100 g of the compound:

Mass of iron in 100 g of compound = 36.76 g

Mass of iron in 100 g of compound = 21.11 g

Mass of iron in 100 g of compound = 42.13 g

Now calculate the number of moles each element:

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Moles of sulfur= $\frac{21.11 g}{32.06 g/mol}=0.658 mol$

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Divide the moles of each element by the smallest number of moles to calculated the ratio of the elements to each other

For Iron element = $\frac{0.658 mol}{0.658 mol}=1$

For sulfur element = $\frac{0.658 mol}{0.658 mol}=1$

For oxygen element = $\frac{2.633 mol}{0.658 mol}=4.001\approx 4$

So, the empirical formula of the compound is $Fe_1S_1O_4$

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1 year ago

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