For the reaction 3KOH+H3PO4⟶K3PO4+3H2O

how many grams of phosphoric acid, H3PO4, are needed to react completely with 89.7 g of potassium hydroxide, KOH?

52.2 g

Explanation:

Step 1: Write the balanced equation

3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O

Step 2: Calculate the moles corresponding to 89.7 g of KOH

The molar mass of KOH is 56.11 g/mol.

89.7 g × 1 mol/56.11 g = 1.60 mol

Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH

The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.

Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.533 mol × 97.99 g/mol = 52.2 g

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:

lol idk man i guess u gotta ask bill nye the science guy