autumnguidry1622
22.07.2019 •
Chemistry
How many milliliters of 0.564 m hcl are required to react with 6.03 grams of caco3 ?
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Ответ:
CaCO3 + 2 HCl = CO2 + CaCl2 + H2O
Find how many moles of CaCO3 you are dealing with
6.03 g x 1 mol / 100 g = .0603 moles
However, you need twice as much HCl
.0603 x 2 moles / .564 M = .21 L
Ответ:
With butane (C 4 H 10), you can again balance the carbons and hydrogens as you write the equation down. Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" O 2 molecules on the left.
Explanation: