Consider the following reaction equation: CaCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) → CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\). What mass of CaCI\(_2\) would be obtained when 25.0g of CaCO\(_{3(s)}\) is reacted with excess HCI\(_{(aq)}\)? [CaCO\(_3 = 100; CaCI\(_2\) = 111]
- 4.00g
- 4.44 g
- 18.9 g
-
27.8 g
The correct answer is: D
Explanation
CaCO\(_3\) + 2HCI → CaCI\(_2\) + H\(_2\)O + CO\(_2\)
100g CaCO\(_3\) will produce 111g CaCI\(_2\)
25g CaCO\(_3\) will produce x ,
x = \(\frac{25 x 111}{100}\)
= 27.75 \(\approxeq\) 27.8g