Na + Br2 --> NaBr Calculate how many grams of NaBr may be produced from 5.00 grams of solid sodium metal when combined with unlimited Bromine gas.

22.39g of NaBr

Explanation:

First let us write a balanced equation for the reaction. This is illustrated below:

2Na + Br2 —> 2NaBr

Molar Mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar Mass of NaBr = 23 + 80 = 103g/mol

Mass of NaBr from the balanced equation = 2 x 103 = 206g

From the equation,

46g of Na produced 206g NaBr.

Therefore, 5g of Na will produce = (5 x 206)/46 = 22.39g of NaBr

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

14,16 18 reacts with group