Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
Answer:
element
A pure substance is matter that is uniform throughout and has consistent properties. All matter is made up of very small particles called atoms. Atoms are the building blocks of all matter. When a substance contains only one type of atom, it is called an element.
Answer: 0.13mol
Explanation:Please see attachment for explanation
<span>1 fm = 1.0 × 10-15 meters
</span>so
0.610<span> fm x (1.0x10^-15 m / 1 fm)
= 6.1x10^-16 m</span>
so the formula is
λ = h / mu
so,
as we know mass value so by putting
<span>m = (6.626*10^-34)/(6.1*10^-16*199)
=</span>(6.626*10^-34)/(1213.96*<span>10^-16)
</span><span> = 5.45 x 10^ -21
hope thats an answer for you</span>
