Peter M. answered • 10d

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__3__ Cu + __8__ HNO_{3}** --> 3 Cu(NO**

_{3}

**)**

_{2}

**+**

__2__NO +__4__H_{2}

**O**

We want to calculate the number of moles of water produced when reacting with 25 moles of HNO_{3}. In order to do this, we can use the mole ratio that is given to us by the balanced chemical equation. These are the coefficients in front of each specie in the equation. The only thing to take into account is that units that are diagonal from each other when multiplying cancel.

25 mol HNO_{3} * (4 mol H_{2}O / 8 mol HNO_{3}) = 12.5 mol H_{2}O

__2__ NaClO_{3}** ---> 2 NaCl + 3 O**

_{2}

The same explanation applies. If we want to make 44 moles of O_{2}, then we just start with that and use the ratio, cancelling along the diagonal, and multiplying to get our answer.

44 mol O_{2} * (2 mol NaClO_{3} / 3 mol O_{2}) = 29.33 mol NaClO_{3}