Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
<span>Among the choices provided, the answer is letter C, CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s) since Iron(Fe) is higher up on the action graph and is, along these lines, more responsive than Copper(Cu). Because of this, Iron can supplant Copper in that substance equation. </span>
Answer:
16.27 g of CaCO3 are produced upon reaction of 45 g CaCN2 and 45 g of H2O.
Explanation:
Ca(CN)2 + 3H2O → CaCO3 + 2 NH3
First of all, let's find out the limiting reactant.
Molar mass Ca(CN)2.
Molar mass H2O: 18 g/m
Moles of Ca(CN)2: mass / molar mass
45 g / 92.08 g/m = 0.488 moles
Moles of H2O: mass / molar mass
45g / 18g/m = 2.50 moles
This is my rule of three
1 mol of Ca(CN)2 needs 3 moles of H2O
2.5 moles of Ca(CN)2 needs (2.5 . 3) / 1 = 7.5 moles
I need 7.5 moles of water, but I only have 0.488. Obviously water is the limiting reactant; now we can work on it.
3 moles of water __ makes __ 1 mol of CaCO3
0.488 moles of water __ makes ___ (0.488 . 1) / 3 = 0.163 moles
Molar mass CaCO3 = 100.08 g/m
Molar mass . moles = mass
100.08 g/m . 0.163 moles = 16.27 g
C2H6 (g) + O2 (g). --> CO2(g)+H2O (g)
2C2H6 (g) + 7O2 (g). --> 4CO2(g)+6H2O (g)
(8.45 x 10^23)/(6.022x10^23) = 1.403 moles H2O
1.403 moles H2O x (4molCO2/6molH2O)= .935 moles CO2
.935 moles CO2 x (6.022 x10^23)= 5.63 x 10^23 molecules CO2
