Question

Suppose two chemical reactions are linked together in a way that the O2 produced in the first reaction goes on to react completely with Mg to form MgO in the second reaction. Reaction one: 2 KClO3 → 3 O2 + 2 KCl Reaction two: 2 Mg + O2 → 2 MgO If you start with 4 moles of KClO3, how many moles of MgO could eventually form

Answer 1

Answer:

Numbe of mole of MgO form=12

Explanation:

First calculate the number of mole of $O_2$ produced from 4 mole of  $KCLO_3$

Balance the first reaction:

$2KClO_3 \rightarrow 3 O_2 + 2 KCl$

from the above balanced reaction it is clearly that,

by unitry method,

2 mole of $KCLO_3$ produced 3 mole of $O_2$

1 mole of $KCLO_3$ produced 1.5 mole of $O_2$

from 4 mole of $KCLO_3$  $4\times 1.5$ mole of $O_2$ produced

hence we have 6 mole of $O_2$ and this total oxygen will react with Mg

Balance the second reaction:

$2Mg + O_2 \rightarrow 2 MgO$

since produced oxygen in first reaction is reacted completely with Mg means Mg is given in excess quantity,

From the second balanced reacion it is clearly that,

1 mole of $O_2$ produced 2 Mole of MgO

hence 6 mole of $O_2$ will produce 12 mole of MgO

Numbe of mole of MgO form=12