P4O10 + 6H20 → 4H3PO4 If 30 g of H2O reacts with 40g of P4O10, what is the limiting reactant? How much H3PO4 can be formed?

Question

P4O10 + 6H20 → 4H3PO4

If 30 g of H2O reacts with 40g of P4O10, what is the limiting reactant? How much H3PO4 can be formed?

in progress 0
Hồng Cúc 5 months 2021-09-05T14:00:46+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-09-05T14:02:04+00:00

    Answer:  P_4O_{10} is the limiting reagent

    66.6 g of H_3PO_4 will be formed.

    Explanation:

    To calculate the moles :

    \text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}  \text{Moles of} P_4O_{10}=\frac{40g}{284g/mol}=0.14moles

    \text{Moles of} H_2O=\frac{30g}{18g/mol}=1.7moles

    P_4O_{10}+6H_2O\rightarrow 4H_3PO_4  

    According to stoichiometry :

    1 mole of P_4O_{10} require 6 moles of H_2O

    Thus 0.14 moles of P_4O_{10} will require=\frac{6}{1}\times 0.14=0.84moles  of H_2O

    Thus P_4O_{10} is the limiting reagent as it limits the formation of product and H_2O is the excess reagent.

    As 1 mole of P_4O_{10} give = 4 moles of H_3PO_4

    Thus 0.17 moles of P_4O_{10} give =\frac{4}{1}\times 0.17=0.68moles  of H_3PO_4

    Mass of H_3PO_4=moles\times {\text {Molar mass}}=0.68moles\times 98g/mol=66.6g

    Thus 66.6 g of H_3PO_4 will be formed.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )