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

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

Chemistry Hồng Cúc 3 years 2021-09-05T14:00:46+00:00 2021-09-05T14:00:46+00:00 1 Answers 5 views 0

Answers ( )

Name*

E-Mail*

Website

Featured image

Browse

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

Answer*

thuthuy · Answer 1 · 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.