Question

For the following reaction, 40.2 grams of sulfuric acid are allowed to react with 27.9 grams of calcium hydroxide. sulfuric acid (aq) + calcium hydroxide (s) calcium sulfate (s) + water (l) What is the maximum amount of calcium sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group6 more group attempts remaining

Answer 1

Answer: a) The maximum amount of calcium sulfate that can be formed is 112.9 grams

b) The formula for the limiting reagent is
`Ca(OH)_2`

c) 0.033 moles of excess reagent
`(H_2SO_4)` are left unreacted.

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of
`H_2SO_4`

`\text{Number of moles}=(40.2g)/(98g/mol)=0.410moles`

b) moles of
`Ca(OH)_2`

`\text{Number of moles}=(27.9g)/(74g/mol)=0.377moles`

`H_2SO_4(aq)+Ca(OH)_2(aq)\rightarrow CaSO_4(s)+2H_2O(l)`

According to stoichiometry :

1 moles of
`Ca(OH)_2(` require 1 mole of
`H_2SO_4`

Thus 0.377 moles of
`Ca(OH)_2/tex] will require=[tex](1)/(1)* 0.377=0.377moles` of
`H_2SO_4`

Thus
`Ca(OH)_2` is the limiting reagent as it limits the formation of product and
`H_2SO_4` is the excess reagent as (0.410-0.377)= 0.033 moles of
`H_2SO_4` are left unreacted.

As 1 mole of
`Ca(OH)_2` give = 1 mole of
`CaSO_4`

Thus 0.83 moles of
`Ca(OH)_2` give =
`(1)/(1)* 0.83=0.83moles` of
`CaSO_4`

Mass of
`CaSO_4=moles* {\text {Molar mass}}=0.83moles* 136g/mol=112.9g`

Thus 112.9 g of
`CaSO_4` will be produced from the given masses of both reactants.