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Natasha2012 [34]
3 years ago
10

Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul

ate the moles in 900. g of hydrogen gas? Set up the math. But DONT DO ANY OF IT. Just leave your answer as a math expression.
Chemistry
2 answers:
FromTheMoon [43]3 years ago
8 0

Answer:

n = \frac{900}{0.090*22.4}

Explanation:

If 1 mole of hydrogen gas occupies 22.4 L, then, the number of moles of hydrogen gas (n) can be found by the volume (V):

1 mole ----- 22.4 L

n ------ V

By a simple direct three rule:

22.4n = V

n = V/22.4

But the volume is mass(m) divided by the density (d)

V = m/d

So

n = \frac{m}{22.4*d}

For a mass of 900 g and a density of 0.090 g/L:

n = \frac{900}{0.090*22.4}

BlackZzzverrR [31]3 years ago
6 0

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming that all caculations are at normal pressure and -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

\rho is the density of hydrogen

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When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

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=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

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mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

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