## Answers

1 Now, The given reaction 8 - 35i (8) + 2N2 (9) siz N4 (S) from the reaction , we get! 3 moles of o reacts with 2 sides of Na i mide si a 28g of so - 3 mode 8 = (3x2878 of so = 899 s similarly, mole Na 2 (2 x 1439 N2 = 289 N2.. 8o 2 mode No = 2(28g) - 56g of No ". 3 sole и s° huouk uf, 2 мө{ы и , = 849 of 8o reacts with 56 g of N.

1 ig si reacts with 56 g of N2 - 0.66g of N2 Then, 2g se reacts with & 0.6682 g of No ( 2 1.32 g of No But the mass of No provided is 1.50g, which is excess, so the so acts as a criting reactant in this reaction and the mass of so N produced will depend on the most of si? Again, from reaction, we get a 3 mole so prodnes mide signa l Now, 3 mole s = .89g si (as calculated before) & I mole sizOf 2 (28 x3 + 1484) g siz Na. s 140 g of size Thus, 84g of se prodenes 140g of S3 N4.

Then, 200g of so will produce 140 x 2.00 g Signa - 3.33g of signa Thus, the mass of Sin Na produced by 2.00 g of so and 1.50 g of Na is 3.33g.

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