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Question
From the equation
\[\ce{3Cu + 8HNO3 -> 3Cu(NO3)2 + 4H2O + 2NO}\]
(Atomic mass Cu = 64, H = 1, N = 14,O = 16)
Calculate the mass of copper needed to react with 63g of HNO3.
Solution
Mol. Mass of 8HNO3 = 8 × 63 = 504 g
For 504 g HNO3, Cu required is = 192 g
So, for 63g HNO3Cu required = `(192 xx 63)/504`
= 24g
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