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

P4O10 + 6H20 → 4H3PO4

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

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1. thuthuy

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.