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3 years ago

47.9 ml hrdrogen is collected at 26° Celsius and 718 torr. Find the volume occupied at STP

Chemistry

1 answer:

Temka [501]3 years ago

4 0

Answer:

41.45 mL

Explanation:

Applying the general gas equation,

PV/T = P'V'/T'............... Equation 1

Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.

make V' the subject of the equation

V' = PVT'/TP'................ Equation 2

Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²

Substitute these values into equation 2

V' = ( 95725.196×0.0479×273)/(299×101000)

V' = 0.04145 dm³

V' = 41.45 mL

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Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0

grandymaker [24]

The given question is incomplete. The complete question is:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced

Answer: 60.0 g of glucose

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

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b) moles of $H_2O$

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$6CO_2+6H_2O\rightarrow C_{6}H_{12}O_6+6O_2$

According to stoichiometry :

6 moles of $CO_2$ require = 6 moles of $H_2O$

Thus 2.0 moles of $CO_2$ require= $\frac{6}{6}\times 2.0=2.0moles$ of $H_2O$

Thus $CO_2$ is the limiting reagent as it limits the formation of product.

As 6 moles of $CO_2$ give = 1 moles of glucose

Thus 2.0 moles of $CO_2$ give = $\frac{1}{6}\times 2.0=0.33moles$ of glucose

Mass of glucose = $moles\times {\text {Molar mass}}=0.33moles\times 180g/mol=60g$

Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water

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