Question

What's the answer of this question?(Remember the correct number of significant digits )

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaoH) to produce aqueous sodium bromide (NaBr) and liquid water (H,0) Suppose 57. g of hydrobromic acid is mixed with 49.1 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits

Answer 1

Molar mass of HBr,

MM = 1*MM(H) + 1*MM(Br)

= 1*1.008 + 1*79.9

= 80.908 g/mol

mass(HBr)= 57.0 g

number of mol of HBr,

n = mass of HBr/molar mass of HBr

=(57.0 g)/(80.908 g/mol)

= 0.7045 mol

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

mass(NaOH)= 49.1 g

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(49.1 g)/(39.998 g/mol)

= 1.228 mol

Balanced chemical equation is:

HBr + NaOH ---> H2O + NaBr

1 mol of HBr reacts with 1 mol of NaOH

for 0.704504 mol of HBr, 0.704504 mol of NaOH is required

But we have 1.227561 mol of NaOH

so, HBr is limiting reagent

we will use HBr in further calculation

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (1/1)* moles of HBr

= (1/1)*0.704504

= 0.704504 mol

mass of H2O = number of mol * molar mass

= 0.7045*18.02

= 12.69 g

According to balanced equation

mol of NaOH formed = (1/1)* moles of HBr

= (1/1)*0.704504

= 0.704504 mol

mol of NaOH remaining = mol initially present - mol reacted

mol of NaOH remaining = 1.227561 - 0.704504

mol of NaOH remaining = 0.523057 mol

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

mass of NaOH,

m = number of mol * molar mass

= 0.5231 mol * 39.998 g/mol

= 20.9 g

Answer: 20.9 g