Answer:
Option D is correct.
H₂O + CO₂ → H₂CO₃
Explanation:
First of all we will get to know what law of conservation of mass states.
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Example:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.
The answer is 6.1*10^-3 atm.
The pictures and explanations are there.
Answer:
Explanation:
We have in this question the equilibrium
X ( g ) + Y ( g ) ⇆ Z ( g )
With the equilibrium contant Kp = pZ/(pX x pY)
The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional
pV = nRT ⇒ p = nRT/V and n/V is molarity.
Therefore we can calculate the reaction quotient Q
Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2
Since Qp is greater than Kp the system proceeds from right to left.
We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce 
g of (NH₄)₂S
25g of NH₃ will produce 
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
Answer:
<span>23.6
g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from
15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because
it gets used up and only makes 10.7 g of CO2. </span>
Explanation:
1) Balanced chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O
3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol
4) Convert the reactant masses to number of moles, using the formula
number of moles = mass in grams / molar mass
CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />
O₂: 15.6 g / 32.0 g/mol = 0.4875 moles
5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂
Which is what the first part of the answer says.
6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.
Which is what the second part of the answer says.
7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.
Which is what the chosen answer says.
8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.
