Answer:
14.5 g silver
Explanation:
This is a problem using the stoichiometry of the reaction. First thing we need is the balanced equation:
Zn + 2 AgNO3 ----------------------- 2 Ag + Zn(NO3)2
We know that 14.6 g of Zn did not reacted, then we can calculate the amount of Zn reacted and do the calculation given the above reaction.
amount Zn reacted: 19.0 -14.6 g Zn = 4.4 g Zn
atomic weight of Zn: 65.37 g/mol
mol Zn reacted: 4.4 g Zn x ( 1 mol Zn/ 65.37 g Zn) = 0.067 mol Zn
We know from the balanced equation that moles of Ag are produced from 1 mol Zn therefore the mol of Ag produced are:
0.067 mol Zn x 2 mol Ag/ 1mol Zn = 0.135 mol Ag
and the mass of silver then will be given by multiplying by the atomic weight of silver:
0.135 mol Ag x 107.9 g/mol = 14.5 g Ag
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
---------------
C = n/V
n = C×V
n = 2.1×0,075
n = 0,1575 mol
--------
mKCl: 39+35.5 = 74,5 g/mol
74,5g --------- 1 mol
Xg ------------- 0,1575 mol
X = 74,5×0,1575
X = 11,73375g KCl
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