A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m
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Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.
The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.
Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles
We can obtain the mass of 3-methyl-1-butanol from its density.
Mass = density × volume
Density of 3-methyl-1-butanol = 0.810 g/mL
Volume of 3-methyl-1-butanol = 9 mL
Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL
Mass of 3-methyl-1-butanol = 7.29 g
Number of moles of 3-methyl-1-butanol = mass/molar mass = 7.29 g/88 g/mol = 0.083 moles
Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol
Mass of product 1-bromo-3-methylbutane = number of moles × molar mass
Molar mass of 1-bromo-3-methylbutane = 151 g/mol
Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol
= 12.53 g
Recall that % yield = actual yield/theoretical yield × 100
Actual yield of product = 6.48 g
Theoretical yield = 12.53 g
% yield = 6.48 g/12.53 g × 100
% yield = 51.7%
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Answer:
0.15g
Explanation:
Given parameters:
Number of molecules of water = 1.2 x 10²¹ molecules
Unknown:
Mass of SnO₂ = ?
Solution:
To solve this problem, we have to work from the known to the unknown specie;
SnO₂ + 2H₂ → Sn + 2H₂O
Ensure that the equation given is balanced;
Now,
the known species is water;
6.02 x 10²³ molecules of water = 1 mole
1.2 x 10²¹ molecules of water = = 0.2 x 10⁻²moles
Number of moles of water = 0.002moles
From the balanced chemical equation:
2 mole of water is produced from 1 mole of SnO₂
0.002 moles of water will be produced from = 0.001moles
To find the mass;
Mass = number of moles x molar mass
Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol
Mass = 0.001 x 150.7 = 0.15g