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3 years ago

Use the periodic table to determine how many grams of oxygen would be required to react completely with 859.0 g C2H2?

2 answers:

6 0

(a) The balanced equation shows that 2 moles of water result from 5 moles of
oxygen that reacts, so that is 0.4 times as many moles of water as O2. So if we
have 2.50 moles of oxygen reacting, we produce 0.4 times that amount, or
1.0 mole of water.
(b) The molecular weight of acetylene is 26 and that of O2 is 32, so we can set
up the proportion: 2.25 gm. / 52 = x gm. / 160
Then x grams = (2.25)(160) / 52 = 6.923 grams of O2 are required.
(c) The balanced equation shows that twice as many moles of CO2 are produced
as we have acetylene reacting. If 78.0 grams of acetylene react, that is 3 moles
so we produce twice that, or 6 moles of CO2. The molecular weight of CO2 is 44,
so we have 44 times 6, or 264 grams of carbon dioxide produced.
(d) If we collect 186 grams of CO2, the percentage yield is:
186/264 = 0.7045, or 70.45 percent yield.
Whew! Hope this answers all parts of your question!

Vsevolod [243]3 years ago

6 0

The grams of oxygen that is required to react completely with 859.0 g C₂H₂ is 2643.2 grams

calculation

Step 1: write the equation for reaction between C₂H₂ and O₂

that is 2C₂H₂ +5O₂ → 4 CO₂ +2H₂O

Step 2 : find the moles of C₂H₂

moles = mass/molar mass

From periodic table the molar mass of C₂H₂ = (12 x2) +( 1 x2) = 26 g/mol

moles= 859.0 g /26 g/mol =33.04 moles

step 3 : use the moles ratio to determine the moles of O₂

C₂H₂:O₂ is 2:5 therefore the moles of O₂ =33.04 x5/2 = 82.6 moles

Step 4: find mass of O2

mass = moles x molar mass

from periodic table the molar mass of O₂ = 16 x2 = 32 g/mol

mass = 82.6 moles x 32 g/mol =2643.2 grams

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Answer:

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Answer:

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