Question

How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure?
Ca + Cl2 ---> CaCl2

Answer 1

please help!

Answer 2

You have 1.40 moles of Ca, so it consumes 1.40 moles of Cl2. at STP, 1 mole of any gas occupies 22.4 L. So multiply 1.40 mol x 22.4 L/mol = 31.36 L of Cl2 gas.

Answer 3

Thanks!

Answer 4

Can you help me out with another one please?

Answer 5

sure.

Answer 6

A laboratory experiment requires 2.25 L of a 1.0 M solution of phosphoric acid (H3PO4), but the only available H3PO4 is a 9.0 M stock solution. How could you prepare the solution needed for the lab experiment?

Answer 7

Well, use the formula (M1)(V1)=(M2)(V2). You know the required molarity and volume (M1 and V1, respectively), and the initial molarity (M1). You are solving for V1, the volume of the stock solution that you dilute to 2.25 L. V1=(M2)(V2)/(M1), so V1=(1.0)(2.25)/(9.0). By this, V1=0.25 L. So you take 0.25 L of stock solution, add it to 2.00 L of distilled water, and you get your 2.25 L of 1.0 M solution.

Answer 8

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Answer 9

You're very welcome.