Question

Help me to solve this question please?

Answer 1

Convert mass to moles, realize that ratio of ammonia to carbon dioxide is 2:1 so ammonia will be the limiting reactant. Since 2 moles of ammonia react with one mole of `CO_2`, (0.03742/2 = 0.1871) moles of `CO_2` can be used, and 0.0724 moles of `CO_2` will be left over.

Explanation:

`((6.372 g) )/ (17.03(g/(mol)))` `= 0.3742mol NH_3`

`((11.42 g)) /(44.01(g/(mol))` `= 0.02595 mol CO_2`
Since 2 moles of ammonia react with one mole of `CO_2`, (0.03742/2 = 0.1871) moles of `CO_2` can be used, and 0.0724 moles of `CO_2` will be left over.

`(0.0724 mol` `CO_2`)`(44.01g/(mol)) = 3.19 g` left over `CO_2`

One mole of urea is formed for each mole of `CO_2` consumed, so 0.1872 moles of urea will be produced.
`(0.1872 (mol urea))(60.06 (g/(mol)))=11.24g` urea produced