Answer:
960g
Explanation:
[tex]C_{12}H_{22}O_{11}+ 3O_{2} \longrightarrow 2H_{3}C_{6}H_{5}O_{7} + 3H_{2}O[/tex]
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
[tex]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}[/tex]
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 [tex]H_{3}C_{6}H_{5}O_{7}[/tex]
[tex]H= 1 g/mol\\C= 12 g/mol\\ O=16g/mol[/tex]
[tex][tex]H_{3}C_{6}H_{5}O_{7}= (8)1g/mol+6(12g/mol)+ 7 (16g/mol)=192 g/mol[/tex][/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 [tex]H_{3}C_{6}H_{5}O_{7}[/tex] are in 5 mol.
[tex]1mol \longrightarrow 192 g \\ 5 mol \longrightarrow x\\ x=\frac{5mol(191g)}{1mol} = 960g[/tex]
