Question

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

P2O3+3H20=2H3PO3


Thanks

Answer 1

To determine the grams of H3PO3 produced from 93.2 g of P2O3, you calculate moles of P2O3, use the balanced equation to find moles of H3PO3, and convert that to grams, yielding approximately 137.55 g H3PO3.

Step-by-step explanation:

To calculate the amount of H3PO3 produced from the complete reaction of 93.2 g P2O3, we must follow these steps:

Molar masses: P2O3 (110.97 g/mol), H3PO3 (81.99 g/mol)

Step 1: Calculate moles of P2O3

93.2 g P2O3 x (1 mol P2O3 / 110.97 g P2O3) = 0.839 moles P2O3

Step 2: Using the balanced equation, 1 mole of P2O3 produces 2 moles of H3PO3. Therefore:

0.839 moles P2O3 x (2 moles H3PO3 / 1 mole P2O3) = 1.678 moles H3PO3

Step 3: Convert moles of H3PO3 to grams:

1.678 moles H3PO3 x (81.99 g H3PO3 / 1 mol H3PO3) = 137.55 g H3PO3

Therefore, 93.2 g P2O3 would produce approximately 137.55 g of H3PO3 when the reaction goes to completion.

Answer 2

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