39.27
approx 2.4354g
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
28g Fe
This question is not specific and can therefore not be answered accurately.
The mass of ammonia is 147,5 g.
6.08
.2M x V FeCl3=moles FeCl3 x 1mole Fe2S3/2mole FeCl3=moles of Fe2S3 x mm of Fe2S3/1 mole Fe2S3= g Fe2S3 x .65% yield. 2.75g Fe2S3/ .65= 4.23g Fe2S3/ 207.91= .02035 x 2mole FeCl3=.0407 moles FeCl3/ .2M FeCl3= .2035 L x 1000= 203.5 ml
there would be 50 grams of ammonia will be formed
384.5g
approx 2.4354g
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
The answer is: approx. 327 g.
25
28g Fe
This question is not specific and can therefore not be answered accurately.
how many grams of oxygen are consumed when 19.4g of carbon dioxide is formed during the combustion of C7H16
There is no mass loss (nor gain) in state change, so there would be 100 grams of ice formed.