Chemistry, asked by abeer31, 2 months ago

A solid mixture weighing 5.00 g containing lead nitrate and sodium nitrate was heated below 600°C
until the mass of the residue was constant. If the loss of mass is 30 %, find the mass of lead nitrate
and sodium nitrate in mixture. (At. wt. of Pb = 207, Na = 23, N = 14,0 = 16)
Given : 2Pb(NO3),(s) - $ 2PbO(s) + 4NO2(g) + O2(g) and
2NaNO3(s) - A 2NaNO,(s) + O2(9)
Let, wt. of Pb(NO3)2 in mixture
wt. of NaNO3 (5 - x) g
662 g of Pb(NO3)2 will give residue = 446
= X​

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Answered by bhat0709
1

Answer:

A solid mixture weighing 5.00 g containing lead nitrate and sodium nitrate was heated below 600°C

until the mass of the residue was constant. If the loss of mass is 30 %, find the mass of lead nitrate

and sodium nitrate in mixture. (At. wt. of Pb = 207, Na = 23, N = 14,0 = 16)

Given : 2Pb(NO3),(s) - $ 2PbO(s) + 4NO2(g) + O2(g) and

2NaNO3(s) - A 2NaNO,(s) + O2(9)

Let, wt. of Pb(NO3)2 in mixture

wt. of NaNO3 (5 - x) g

662 g of Pb(NO3)2 will give residue = 446

= X

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