How many grams of water are needed to react with 485g of cac2

273 g  

Step-by-step explanation:

a) Balanced equation

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:    64.10   18.02

          CaC₂ + 2H₂O ⟶ Ca(OH)₂ + 2C₂H₂

m/g:    485

(b) Mass of H₂O

(i) Calculate the moles of CaC₂

n = 485 g CaC₂ × (1 mol CaC₂/64.10 g CaC₂)

  = 7.566 mol CaC₂  

(ii) Calculate the moles of H₂O

The molar ratio is (2 mol H₂O /1 mol CaC₂)

n = 2.00 mol CaC₂ × (7.566 mol H₂O /1 mol CaC₂)

  = 15.13 mol H₂O

(iii) Calculate the mass of H₂O

m = 165.16 mol H₂O × (18.02 g H₂O /1 mol H₂O)

    = 273 g H₂O

The reaction requires 273 g H₂O.

first one is the correct answer

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g