1: 34 g Mg combines with 50.00 mL of 0.30 M nitric acid acid. The hydrogen produ

Question

1: 34 g Mg combines with 50.00 mL of 0.30 M nitric acid acid. The hydrogen produced was collected over water at 20.0 C with a total pressure of 660.0 torr.    A: Find the volume of the hydrogen gas. (Hint: Find the limiting reactant)

2: Plants produce sugar (C6H12O6) based on photosynthesis process. They absorb water and minerals from the soil through their roots and carbon dioxide through their leaves.

Write a balanced equation for this process and calculate the volume of carbon dioxide consumed if 50.00 g sugar is produced at 25.0 and 1.00 atm .

3: KClO3 decomposes to produce oxygen gas and KCl. Suppose 2.00 L O2 is be produced at 20.0 C and a total pressure of 740.0 torr. Suppose the oxygen gas is collected over water and the vapor pressure of water is 17.9 torr at 20.0 C. Calculate the mass of the KClO3 used.

Explanation / Answer

1) Mg(s) + 2HCl(aq) ---> MgCl2(aq) + H2(g)

34 g Mg combines with 50.00 mL of 0.30 M nitric acid acid.

moles of Mg = 34/ 24.3050 = 1.398 moles of Mg

no of moles of HCl = concentration * volume = 0.30*0.050 = 0.015 MOLES OF HCL

so 0.015 moles of HCL can reat with 0.0075 moles of Mg so HCL is less comared to Mg

( 0.015 mol HCL) x (1 mol H2 / 2 mol HCL) x (2 g H2/mol) = 0.015 g H2

no of moles of H2 that can be produced = 0.0075 moles of H2

pV = nRT

p = 0.868421 atm

n = 0.0075

T = 273 +20 = 293 K

R = 0.082057 L atm mol-1K-1

V = 0.0075*293*0.082057 / 0.868421 = 207ml of H2 is produced

2)6CO2 + 6H2O ===> C6H12O6 + 6O2

the volume of carbon dioxide consumed if 50.00 g sugar is produced at 25.0 and 1.00 atm

moles of sugar = 50grams of sugar / 180.157 g/mol.=0.2775 moles

6 moles of co2 is required to produce 1 mole of sugar so

1.665 moles of Co2 is required to produce 0.2775 moles of C6H12O6

pV = nRT

p = 1 atm

n = 1.665

T = 273 +25 = 298 K

R = 0.082057 L atm mol-1K-1

V = 1.665*298*0.082057 /1 = 40.71L of Co2 is consumed

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