1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:
We can calculate the mass percent of O in MgO using the following expression.
You can learn more about mass percent here: brainly.com/question/14990953
Answer:
B
Explanation:
its an acidic oxide, it disolves in water to form carbonic acid which is an acid
Answer:
17.09g/L
Explanation:
Density = total mass of elements/ volume
We need to find the mass of each mixture constituents using their molar mass:
mole = mass/molar mass
For Neon (Ne) which contains 0.650mol;
0.650 = mass/20.18
mass = 0.650 × 20.18
mass = 13.12g
For Krypton (Kr) which contains 0.321mol;
0.321 = mass/83.79
mass = 0.321 × 83.79
mass = 26.89g
For Xenon (Xe) which contains 0.190mol;
0.190 = mass/131.3
mass = 0.190 × 131.3
mass = 24.95g
Total mass = 13.12g + 26.89g + 24.95g = 64.96g
Density = total mass / volume
Density = 64.96g / 3.80L
Density of the mixture = 17.09g/L
The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %
