<span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water.
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100.
One mole of Calcium Nitrate produces one mole of Calcium Carbonate.
i.e. 164 gms will produce 100gms of precipitate
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>
Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
The redox reaction is

Here
Calcium undergoes reduction, and acts as cathode
Lithium undergoes oxidation and acts as anode
The reduction potential of calcium is -2.87 V
The reduction potential of lithium is - -3.05 V
We know that
Ecell = Ecathode - Eanode
Ecell = -2.87 - (-3.05) = 0.18 V
<span>NO2 weighs 46.005 grams per mol. There are 6.02x10^23 molecules in a mol. In the given sample of 189.5 grams, there are 4.12 mols. This means that there are 2.48x10^24 molecules of NO2 in the given sample.</span>