Question

Citric acid (h3c6h5o7) is a product of the fermentation of sucrose (c12h22o11) in air. determine the mass of citric acid produced when 2.50 mol c12h22o11 is used. the balanced equation is: c12h22o11 + 3 o2 → 2 h3c6h5o7 + 3 h2o

Answer 1

2.50 x 2/1 = 5 mol of Citric Acid
5 x (3+72+5+112) = 960g of Citric Acid


Answer: 960g of Citric Acid

Answer 2

Answer:

960g

Explanation:

$C_{12}H_{22}O_{11}+ 3O_{2} \longrightarrow 2H_{3}C_{6}H_{5}O_{7} + 3H_{2}O$

We use stoichiometric relations to solve it.

we know that 1 mol of sucrose produces 2 mol of citric acid (we know it by the coefficients of balancing of the equation)

Now how many moles of citric acid  are produced when2.50 mol of sucrose are used

$1mol C_{12}H_{22}O_{11}\longrightarrow 2 mol H_{3}C_{6}H_{5}O_{7}}\\ 2.50 mol C_{12}H_{22}O_{11}\longrightarrow x}\\ x=\frac{2.50.(2)} 1}{1} =5 mol H_{3}C_{6}H_{5}O_{7}$

We have to find the molar mass of the compound.

We must multiply the atomic mass of each element of the compound by the number of atoms of each one

molar mass $H_{3}C_{6}H_{5}O_{7}$

$H= 1 g/mol\\C= 12 g/mol\\ O=16g/mol$

$[tex]H_{3}C_{6}H_{5}O_{7}= (8)1g/mol+6(12g/mol)+ 7 (16g/mol)=192 g/mol$[/tex]

We know that in 1 mol of Citric acid it has a mass of 192g.

Now we can find out how many grams of  $H_{3}C_{6}H_{5}O_{7}$ are in 5 mol.

$1mol \longrightarrow 192 g \\ 5 mol \longrightarrow x\\ x=\frac{5mol(191g)}{1mol} = 960g$