Answer:
A. Nickel (Ni)
B. 60.28g
Explanation:
A. The balanced equation for the reaction is given below:
2AgNO3 + Ni —> 2Ag + Ni(NO3)2
Next, let us calculate the masses of AgNO3 and Ni that reacted from the balanced equation.
This is illustrated below:
Molar Mass of AgNO3 = 108 + 14 + (16x3) = 108 + 14 +48 = 170g/mol
Mass of AgNO3 from the balanced equation = 2 x 170 = 340g
Molar Mass of Ni = 59g/mol
To obtain the excess reactant, let consider the fact that all the mass sample of AgNO3 is used up in the reaction and see if there will be left over for Ni. If there is no left over then we'll consider the other way round.
From the balanced equation above,
340g of AgNO3 reacted with 59g of Ni.
Therefore, 112g of AgNO3 will react with = (112 x 59)/340 = 19.44g of Ni
Now let us check if there are left over for Ni. This is illustrated below:
Mass of Ni given from the question = 22.9g
Mass of Ni that reacted = 19.44g
Left over Mass of Ni = Mass of Ni from the question - Mass of Ni that reacted
Left over Mass of Ni = 22.9 - 19.44
Left over Mass of Ni = 3.46g
Since there are left over for Ni, therefore nickel (Ni) is in excess and AgNO3 is the limiting reactant.
B. To obtain the mass of nickel(II) nitrate, Ni(NO3)2, formed, the limiting reactant (AgNO3) is used.
The equation for the reaction is given below:
2AgNO3 + Ni —> 2Ag + Ni(NO3)2
Molar Mass of Ni(NO3)2 = 59 + 2[14 + (16x3)] = 59 + 2[14 + 48] = 59 + 2[62] = 59 + 124 = 183g/mol
Mass of AgNO3 from the balanced equation = 340g
From the balanced equation above,
340g of AgNO3 produced 183g of Ni(NO3)2.
Therefore, 112g of AgNO3 will produce = (112 x 183)/340 = 60.28g of Ni(NO3)2
From the calculations made above, 60.28g of Ni(NO3)2 is produced from the reaction of 22.9g of Ni and 112g of AgNO3
