Question

For the following reaction, 12.6 grams of sodium are allowed to react with5.96 grams of water.

sodium(s) + water (I) ---> sodium hydroxide (aq) + hydrogen (g)

What is the maximum amount of sodium hydroxide that can be formed? in grams

What is the formula for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? in grams

Answer 1

1)

Molar mass of Na = 22.99 g/mol

mass(Na)= 12.6 g

number of mol of Na,

n = mass of Na/molar mass of Na

=(12.6 g)/(22.99 g/mol)

= 0.5481 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 5.96 g

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(5.96 g)/(18.016 g/mol)

= 0.3308 mol

Balanced chemical equation is:

2 Na + 2 H2O ---> 2 NaOH + H2

2 mol of Na reacts with 2 mol of H2O

for 0.548064 mol of Na, 0.548064 mol of H2O is required

But we have 0.330817 mol of H2O

so, H2O is limiting reagent

we will use H2O in further calculation

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

According to balanced equation

mol of NaOH formed = (2/2)* moles of H2O

= (2/2)*0.330817

= 0.330817 mol

mass of NaOH = number of mol * molar mass

= 0.3308*40

= 13.23 g

Answer: 12.2 g

2)

H2O is limiting reagent

3)

According to balanced equation

mol of Na reacted = (2/2)* moles of H2O

= (2/2)*0.330817

= 0.330817 mol

mol of Na remaining = mol initially present - mol reacted

mol of Na remaining = 0.548064 - 0.330817

mol of Na remaining = 0.217247 mol

Molar mass of Na = 22.99 g/mol

mass of Na,

m = number of mol * molar mass

= 0.2172 mol * 22.99 g/mol

= 5.00 g

Answer: 5.00 g