Chemical reactions/stoichiometry help!!

[bouch037]

Help with chemical reaction questions:
I'm stuck on these two problems. Any help is appreciated!


If you needed to produce 375 g of P₂O₅, how many g of P₄ should you start with assuming no loss in the reaction.
P₄ + O₂ → P₂O₅


The United States was reported to have released 5.267x10^9 g of CO₂ in 2013. If this CO₂ all reacted with water to form carbonic acid (H₂CO₃), how many grams of water would be needed?
CO₂ + H₂O → H₂CO₃

[ChenBeier]

First develop the correct balanced equation.

P₄ + O₂ => P₂O₅ is not balanced.

Then calculate the moles of the 375 g P₂O₅, this correspond according the equation to how much mole P₄?
Covert back to grams.

For the second example go the same way. The equation is already correct. Convert also here the g to mol and look how many moles H₂O correspond to it. Convert back to gram.

Mole (n) = m/M , m= mass, M = molar mass

[Thiandes]

Thank a lot!

[johnnysmith]

I’m working on some chemical reaction and stoichiometry problems, but I’m still a bit confused on a few steps. I saw bouch037’s post and ChenBeier’s reply, but I’d like to ask about another similar example to get a better grasp.

Suppose I need to produce 250 g of Fe₂O₃ from this reaction:

Fe + O₂ → Fe₂O₃

1. How do I balance this equation?
2. How many grams of Fe would I need to start with, assuming no loss in the reaction?
I know I need to convert to moles and use the ratio from the equation, but I often mess up the calculation part. Could someone explain it step-by-step in detail? Thanks so much, everyone!

[ChenBeier]

You see Fe₂O₃ is made of 2 Fe and 3 O

So simple you can wrote first 2 Fe + 3 O => Fe₂O₃.

Now oxygen exist as O₂ . So you have to multiply everything by 2

4 Fe + 3 O₂ => 2 Fe₂O₃ balanced equation.

For the amount you built the ratio with molar mass.

250 g Fe₂O₃ /( 2*159,69 g/mol) = xg Fe/(4*55,85 g/mol)

Solve for x = 174,87 g Fe

[bevis0405]

To produce 375 g of P₂O₅, you need 163.6 g of P₄.
To react with 5.267 × 10⁹ g of CO₂ to form H₂CO₃, you need 2.155 × 10⁹ g of water.