1) How many milliliters of 0.55 M HCl are needed to react with 6.2 g of CaCO 3 ?

Question

1) How many milliliters of 0.55 M HCl are needed to react with 6.2 g of CaCO3?

- mL

2) How many grams of NaH2PO4 are needed to react with 37.11 mL of 0.285 M NaOH?

- g

3) How many grams of solid barium sulfate form when 16.9 mL of 0.160 M barium chloride reacts with 58.5 mL of 0.055 M sodium sulfate? Aqueous sodium chloride is the other product.
- g

4) to prepare a fertilizer, an engineer dilutes a stock solution of sulfuric acid by adding 22.1 L of 3.00 M acid to enough water to make 500. L. What is the mass of sulfuric acid per milliliter of the diluted solution?
- g per milliliter

Explanation / Answer

molar mass of CaCO3 100.1 , moles of CaCO3 = 6.2/100.1 = 0.062 moles, according to the above equation, 2 moles of HCl is needed for one mole of CaCO3. so moles of HCl will be 0.062 x 2 = 0.124 moles.

now, 0.55 x V/1000 = 0.124 moles; where V is the vol of HCl solution req.

so, V = 225.2 mL.

2.   37.11 mL of 0.285 M NaOH = 37.11 x 0.285 /1000 moles,

for that moles of NaH2PO4 required = 37.11 x 0.285 /1000 x 2, so g of NaH2PO4

= 37.11 x 0.285 x 119.98 /1000 x 2 = 0.634 g.

4. 22.1 L of 3.00 M acid = 22.1 x 3.00 moles present in 500 L. Its mass = 22.1 x 3.00 x 98.1 = 6504 g

so mass per mL = 6504 /5000 x 1000 = 1.3 x 10-3 g = 1.3 mg.

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