Question

2) __H,PO, +

KOH →
K PO + H2O
a- Balance the equation
b- Find the limiting reactant when 2.5 g H,PO, react with 4 g of KOH
C- What is the excess reactant?
d- How much of the excess reactant remained after reaction proceeded?

Answer 1

Step-by-step explanation:

a.H2PO4 + 2KOH = K2PO4 + 2H2O

b. H2PO4 _ 2KOH

1(2) + 31 + 16(4) _ 2(23) + 2(16) + 2(1)

2 + 31 + 64 _ 46 + 32 + 2

97 _ 80

no of moles of H2PO4 = 2.5 ÷ 97

= 0.026mol

no of moles of 2KOH = 4 ÷ 80

= 0.05mol

2KOH is the limiting reactant

c. H2PO4 is the excess reactant

d. 1 mole H2PO4 weighs 2.5g

0.026mol H2PO4 weighs g

(0.026 × 2.5)

H2PO4 weighed 0.065g after reaction