Answer:
D. 15g
Explanation:
The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.
In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.
Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.
Answer:
Sedimentary rock, rock formed at or near Earth's surface by the accumulation and lithification of sediment by the precipitation from solution at normal surface temperatures
Explanation:
Answer:
mass of CO2 = 88 g
Explanation:
It is stated in question that 18.7 g of oxygen remained unreacted which means that carbon is limiting reactant. Thus the amount of carbon dioxide produced depend upon the amount of carbon.
Given data:
mass of carbon = 24 g
total mass of oxygen 82.7 g
mass of unreacted oxygen = 18.7 g
mass of carbon dioxide = ?
Solution:
First of we will calculate the moles of carbon.
number of moles of carbon = mass of carbon / atomic weight
number of moles of carbon = 24 g/ 12 g/mol = 2 mol
Chemical equation:
C + O2 → CO2
now we compare the number of moles of carbon and carbon dioxide,
C : CO2
1 : 1
2 : 2
Now we will calculate the mass of carbon dioxide from moles.
number of moles of CO2 = mass of CO2 / molar mass of CO2
2 mol × 44 g/mol = mass
mass of CO2 = 88 g
Answer:
Explanation:
3
Explanation:
The reaction expression is given as:
Al(OH)₃ + HNO₃ → H₂O + Al(NO₃)₃
To solve this problem, let us assign coefficient a,b,c and d to each specie;
aAl(OH)₃ + bHNO₃ → cH₂O + dAl(NO₃)₃
Conserving Al : a = d
O: 3a + 3b = c + 9d
H: 3a + b = 2c
N: b = 3d
let a = 1 , d = 1, b = 3 , c = 3
Multiply through by 3;
a = 1, b = 3, c = 3 and d = 1
Al(OH)₃ + 3HNO₃ → 3H₂O + Al(NO₃)₃
