Answer:
A. 4Ni(s) + 3O2(g) —> 2Ni2O3(s)
B. Synthesis reaction.
C. 18.31g of O2
Step-by-step explanation:
A. The equation for the reaction.
Ni(s) + O2(g) —> Ni2O3(s)
The above equation can be balance as follow:
There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front Ni2O3 and 3 in front of O2 as shown below:
Ni(s) + 3O2(g) —> 2Ni2O3(s)
There are 4 atoms of Ni on the right side and 1 atom on the left it can be balance by putting 4 in front of Ni as shown below:
4Ni(s) + 3O2(g) —> 2Ni2O3(s)
Now the equation is balanced
B. From the equation, we can see that two reactants combined to form a single product. Therefore, the reaction is termed synthesis reaction.
C. We'll begin by calculating the mass of Ni and O2 that reacted from the balanced equation. This is illustrated below:
4Ni(s) + 3O2(g) —> 2Ni2O3(s)
Molar mass of Ni = 59g/mol
Mass of Ni from the balanced equation = 4 x 59 = 236g
Molar mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 3 x 32 = 96g
Summary:
From the balanced equation above,
236g of Ni reacted with 96g of O2.
Now, we can obtain the mass of O2 needed to react with 45g of Ni as follow:
From the balanced equation above,
236g of Ni reacted with 96g of O2.
Therefore, 45g of Ni will react with = (45 x 96)/236 = 18.31g of O2.
Therefore, 18.31g of O2 is needed to react with 45g of Ni