Question

3.523 x 10^24 molecules of Mg(OH2) will need how many grams of H3PO4 in the reaction below

H=1.01 g/mol
P=30.97 g/mol
O=16.00 g/mol

3 Mg(OH)2 + 2 H3PO4 --------> 1 Mg3 (PO4 )2 + 6 H2O

Answer 1

382.2g of H3PO4 are needed

Step-by-step explanation:

Based on the reaction, 3 moles of Mg(OH)2 react with 2 moles of H3PO4.

To solve this question we must find the moles of Mg(OH)2 using Avogadro's number. Then, using the reaction we can find the moles of H3PO4 and its mass as follows:

Moles Mg(OH)2:

3.523x10²⁴ molecules * (1mol / 6.022x10²³ molecules) = 5.850 moles Mg(OH)2

Moles H3PO4:

5.850 moles Mg(OH)2 * (2 mol H3PO4 / 3 mol Mg(OH)2) = 3.90 moles H3PO4

Mass H3PO4:

Molar mass:

3H = 1.01g/mol*3 = 3.03

P = 30.97g/mol*1 = 30.97

4O = 16.00g/mol*4 = 64.00g/mol

3.03 + 30.97 + 64.00 = 98.00g/mol

3.90 moles H3PO4 * (98.00g / mol) =

382.2g of H3PO4 are needed