S + 6 HNO3 --> H2SO4 + 6 NO2 + 2 H2O How many moles of HNO3 must react to form 14.5 moles of H2O?

43.5 moles of HNO₃.

Explanation:

The balanced equation for the reaction is given below:

S + 6HNO₃ —> H₂SO₄ + 6NO₂ + 2H₂O

From the balanced equation above,

6 moles of HNO₃ reacted to produce 2 moles of H₂O.

Finally, we shall determine the number of mole of HNO₃ required to produce 14.5 moles of H₂O.

This can be obtained as illustrated below:

From the balanced equation above,

6 moles of HNO₃ reacted to produce 2 moles of H₂O.

Therefore, Xmol of HNO₃ will react to produce 14.5 moles of H₂O i.e

Xmol of HNO₃ = (6 × 14.5)/2

Xmol of HNO₃ = 43.5 moles

Therefore, 43.5 moles of HNO₃ is required to produce 14.5 moles of H₂O.

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

it is a good conductor

explanation: