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ss7ja [257]
3 years ago
11

PHEASE HELP ASAP

Chemistry
1 answer:
Veronika [31]3 years ago
6 0

Given the balanced chemical equation for combustion of propane:

C_{3}H_{8}+5O_{2}-->3CO_{2}+4H_{2}O

Moles of propane C_{3}H_{8}=3.6 mol

Propane is the limiting reactant as the other reactant oxygen is said to be present in excess.

Amount of products formed would depend on the moles of limiting reactant.

Mole ratio of water to propane as per the balanced chemical equation

                                   =\frac{4molH_{2}O }{1molC_{3}H_{8}}

Calculating the moles of water produced from 3.6 mol propane:

  3.6molC_{3}H_{8}*\frac{4molH_{2}O }{1molC_{3}H_{8}}

                          = 14.4molH_{2}O

Theoretical yield of water = 14.4 mol

Actual yield = 12 mol

Percent yield = \frac{ActualYield}{TheoreticalYield}*100

                      = \frac{12 mol}{14.4 mol}*100=83.3%

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<span>Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
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