Answer:
5.7 moles of O2
Explanation:
We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:
2KClO3 —> 2KCl + 3O2
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.
Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.
Thus, 5.7 moles of O2 were obtained from the reaction.
The correct option is: CH4 + O2 → CO2 + H2O.
In writing chemical equations, the reactants are usually written to the left of the equation of the reaction while the products are written to the right of the reaction. An arrow pointing in the right direction shows the direction of the reaction. In the question given above, methane and oxygen are the reactants while carbon dioxide and water are the products. The options that states the chemical reaction correctly is option 1.
Answer:
3.8 M
Explanation:
Volume of acid used VA= 57.0 - 37.5 = 19.5 ml
Volume of base used VB= 67.8 - 45.0 = 22.8 ml
Equation of the reaction
2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)
Number of moles of acid NA= 2
Number of moles of base NB= 1
Concentration of acid CA= ???
Concentration of base CB= 1.63 M
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA= 1.63 × 22.8 × 2/ 19.5 × 1
CA= 3.8 M
HENCE THE MOLARITY OF THE ACID IS 3.8 M.
