It seems that you have missed the necessary options for us to answer this question, but anyway here is the answer. William adds two values, following the rules for using significant figures in computations. He should write the sum of these two numbers by using the same number of decimal places as the least precise value. Hope this helps.
Answer:
Explanation:
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Answer:
The percentage yield is 78.2g
Explanation:
Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.
Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)
First we need to calculate the moles of propane
Moles of propane =
g.mol-1
= 0.971 moles
So, moles of CO2 from the moles of propane
1 mole of C3H8(g) = 3 moles of CO2(g)
So, 0.971 moles of C3H8(g) = ?
= 2.913 moles of CO2
So theoretical yield = 2.913 moles
44.0 g/mol
= 128.2 g
So, the actual mass of CO2 = percent yield
theoretical yield / 100 %
= 61.0 %
128.2 g / 100 %
= 78.2 g
the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g
Answer:
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