0

Guest

Courses New

Calculate the amount of calcium oxide required when it reacts with 852 g of P4=O10.

Calculate the amount of calcium oxide required when it reacts with 852 g of P4=O10.

Grade:10

3 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
Sol. The balanced reaction is : 6CaO + P4O10 → 2Ca3 (PO4)2 Moles of P4O10 = 852/284 = 3 Moles of CaO required = 3 × 6 = 18 Mass of CaO required = 18 × 56 = 1008 g
samudra
32 Points
4 years ago
reaction
6CaO+P_{4}O_{10}\rightarrow 2Ca_{3}(PO_{4})_{2}
here,
no. of moles of P4O10=852/284=3
thus,3 moles of CaO is also required
now,
3=M/(6*56)
\RightarrowM=1008g
therefore, 1008g of CaO is required
ankit singh
askIITians Faculty 614 Points
3 years ago
6 moles of CaO reacts with 1 mole of P₄O₁₀ to give 2 moles of Ca₃(PO₄)₂. But there are 3 moles of P₄O₁₀ present in the sample. So, 18 moles of CaO reacts with 3 moles of P₄O₁₀ to give 6 moles of Ca₃(PO₄)₂. Therefore 1008 grams of CaO is required to react with 852 grams of P₄O₁₀.

 

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More