In: Chemistry
balanced reaction: 3Na2CO3 + 2H3PO4 --> 2Na3PO4 +
3H2CO3
1.) How many grams sodium phosphate can be prepared from 43g of
sodium carbonate?
2.) How many grams sodium phosphate can be prepared from 195 mL of
0.50M phosphoric acid?
3.) If 43g of sodium carbonate and 195mL of 0.50M phosphoric acid
are combined which on is the limiting reagent?
4.) how many grams sodium phosphate can be made from reaction
condition in question 3?
5.) if the acid in question 2 is diluted with 500mL of water what
would is its new concentration?
Ans
1. 3 mol of Na2CO3 give 2 mol of NaPO4
1 mol of Na2CO3 give 2/3 mol of NaPO4
106g of Na2CO3 give 2*118/3 g of NaPO4 [ Mol. weight of Na2CO3 is 106 and NaPO4 is 118]
43g of Na2CO3 give (2*118*43)/3*106 = 31.91g of NaPO4
so 31.91g of NaPO4 will produced
2. 2mol of H3PO4 give 2 mol of NaPO4
1 mol of H3PO4 give 1 mol of NaPO4
98 g of H3PO4 give 118 NaPO4
1 g of H3PO4 give 118/98 g of NaPO4
now, in question we have we are using 195ml of 0.50M H3PO4
1000ml of H3PO4 has 0.5 mol of H3PO$
195ml of H3PO4 will have (0.5*195)/1000 = 0.098 mol of H3PO4
so 0.098 mol of H3PO4 will give 0.098 mol of NaPO4
0.098*98 g of H3PO4 will give 0.098*118g of NaPO4
9.6 g of H3PO4 will give 0.098*118g 11.57g of NaPO4
so ans is 195ml 0.5M will give 11.57g of NaPO4
3.
The limiting reagent is the species which completly consumed in the reaction.
43g of Na2CO3 will be 0.405 M
and 195ml of H3PO4 will gave 0.098 M
and in the equstion the ration of H3PO4 and Na2CO3 is 2/3
so we need 2/3 of Na2CO3 for H3PO4 to react
in the queston the (0.405 * 2)/3 = 0.27 not equlat to 0.098 H3PO4 ( less )
so H3PO4 is the limiting reagent.
4. So from question 3, the limitimg reagent is H3PO4, meaning H3PO4 will be completely consumed in the reaction and after that the reaction will be stopped. So, the maximum amount of NaPO4 will be same as of H3PO4 amount.
The ans is 11.57g of NaPO4
5. if the acid is diluted to 500ml the concentartion would be
use this formula
M1 * V1 = M2 * V2
0.5 * 195 = M2 * 500
M2 = 0.195M