Question

Using the balanced equation below, how many grams of h3po3 would be produced from the complete reaction of 93.2 g p2o3?

Answer 1

First  step;  calculate  the  number  of  moles  of  P2O3   reacted
  that  is  mole=mass  divided  by  RFM
RFM  of  P2O3=110
moles  is  therefore=93.2/110=0.847moles
since the  reacting  ratio  of  P2O3 to H3PO3   is  1:2  moles  of  H3PO4  is  therefore =0.847  x2  =1.694moles
mass=moles  x  RFM  of  H3PO3
1.694  x82=138.908g

Answer 2

Answer : The mass of $H_3PO_3$ will be, 137.76 g

Explanation : Given,

Mass of $P_2O_3$ = 93.2 g

Molar mass of $P_2O_3$ = 110 g/mole

Molar mass of $H_3PO_3$ = 82 g/mole

First we have to calculate the moles of $P_2O_3$.

$\text{Moles of }P_2O_3=\frac{\text{Mass of }P_2O_3}{\text{Molar mass of }P_2O_3}=\frac{93.2g}{110g/mole}=0.84moles$

Now we have to calculate the moles of $H_3PO_3$.

The balanced chemical reaction is,

$P_2O_3+3H_2O\rightarrow 2H_3PO_3$

From the balanced reaction we conclude that

As, 1 mole of $P_2O_3$ react to give 2 moles of $H_3PO_3$

So, 0.84 moles of $P_2O_3$ react to give $\frac{2}{1}\times 0.84=1.68$ moles of $H_3PO_3$

Now we have to calculate the mass of $H_3PO_3$.

$\text{Mass of }H_3PO_3=\text{Moles of }H_3PO_3\times \text{Molar mass of }H_3PO_3$

$\text{Mass of }H_3PO_3=(1.68mole)\times (82g/mole)=137.76g$

Therefore, the mass of $H_3PO_3$ will be, 137.76 g