Molar mass of N2 = 28
Moles of N2 = 25 / 28 = 0.89
So, moles of NH3 produce = 2 x 0.89 = 1.78
Note: H2 is in excess. so no need to care about it.
The choices can be found elsewhere and as follows:
a. mass-mass problems
<span>b. mass-volume problems </span>
<span>c. mass-particle problems </span>
<span>d. volume-volume problems
</span>
I believe the correct answer is option D. It is volume-volume problems that does not require the use of molar mass. <span> Here you are dealing with molarities and volumes to determine concentrations. Molar mass is not part of any calculations.</span>
Answer:
Let the mixture is X% by mass of CuSO
4
.5H
2
O and 100 - X % by mass of MgSO
4
.7H
2
O. 5.0 g of mixture will contain 0.05X g CuSO
4
.5H
2
O and 5.0 - 0.05X g MgSO
4
.7H
2
O
The molar masses of CuSO
4
.5H
2
O and MgSO
4
.7H
2
O are 249.7 g/mol and 246.5 g/mol respectively.
The number of moles of CuSO
4
.5H
2
O=
249.7
0.05X
=2.00×10
−4
X moles.
Explanation:
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