Question

Calculate the mass of CuO which can react with 39,2 grams of orthophosphate acid.Please Help!!3CuO+ 2H3PO4 = Cu3(PO4)2 + 3H20

Answer 1

47.73 g CuO

Step-by-step explanation

Given:

Chemical equation: 3CuO+ 2H3PO4 = Cu3(PO4)2 + 3H20

mass of orthophosphate acid (Cu3(PO4)2) = 39.2 g

Required: The mass of CuO

Solution:

`\begin{gathered} 39.2g\text{ H}_3PO_4\text{ x }\frac{1\text{ mol H}_3PO_4}{97,994g\text{ H}_3PO_4}\text{ x }\frac{3\text{ moles CuO}}{2\text{ moles H}_3PO_4}\text{ x }\frac{79,545g\text{ CuO}}{1mole\text{ CuO}} \\ \\ =\text{ 47.73 g CuO} \end{gathered}`

Second method:

Step 1: Find the moles of H3PO4

n = m/M where m is the mass and M is the molar mass of H3PO4

n = 39.2g/97.994g.mol^-1

n = 0.400 mol

Step 2: Use the stoichiometry to find the moles of CuO

The molar ratio between CuO and H3PO4 is 3:2

Therefore the moles of CuO = 0.400 mol x (3/2) = 0.600 mol

Step 3: Find the mass of CuO, now that we have moles

m = n x M m is the mass, n is the moles and M is the molar mass

m = 0.600 mol x 79,545 g/mol

m = 47.73 g